Category Archives: co-ordinate geometry

Cartesian System, Straight Lines: IITJEE Mains: Problem Solving Skills II

September 23, 2017 – 11:08 am

I have a collection of some “random”, yet what I call ‘beautiful” questions in Co-ordinate Geometry. I hope kids preparing for IITJEE Mains or KVPY or ISI Entrance Examination will also like them.

Problem 1:

Given n straight lines and a fixed point O, a straight line is drawn through O meeting lines in the points R_{1}, R_{2}, R_{3}, \ldots, R_{n} and on it a point R is taken such that \frac{n}{OR} = \frac{1}{OR_{1}} + \frac{1}{OR_{2}} + \frac{1}{OR_{3}} + \ldots + \frac{1}{OR_{n}}

Show that the locus of R is a straight line.

Solution 1:

Let equations of the given lines be a_{i}x+b_{i}y+c_{i}=0, i=1,2,\ldots, n, and the point O be the origin (0,0).

Then, the equation of the line through O can be written as \frac{x}{\cos{\theta}} = \frac{y}{\sin{\theta}} = r where \theta is the angle made by the line with the positive direction of x-axis and r is the distance of any point on the line from the origin O.

Let r, r_{1}, r_{2}, \ldots, r_{n} be the distances of the points R, R_{1}, R_{2}, \ldots, R_{n} from O which in turn \Longrightarrow OR=r and OR_{i}=r_{i}, where i=1,2,3 \ldots n.

Then, coordinates of R are (r\cos{\theta}, r\sin{\theta}) and of R_{i} are (r_{i}\cos{\theta},r_{i}\sin{\theta}) where i=1,2,3, \ldots, n.

Since R_{i} lies on a_{i}x+b_{i}y+c_{i}=0, we can say a_{i}r_{i}\cos{\theta}+b_{i}r_{i}\sin{\theta}+c_{i}=0 for i=1,2,3, \ldots, n

\Longrightarrow -\frac{a_{i}}{c_{i}}\cos{\theta} - \frac{b_{i}}{c_{i}}\sin{\theta} = \frac{1}{r_{i}}, for i=1,2,3, \ldots, n

\Longrightarrow \sum_{i=1}^{n}\frac{1}{r_{i}}=-(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})\cos{\theta}-(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})\sin{\theta}

\frac{n}{r}=-(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})\cos{\theta}-(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})\sin{\theta} …as given…

\Longrightarrow (\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})r\cos{\theta}+(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})r\sin{\theta} + n=0

Hence, the locus of R is (\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})x+(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})y+n=0 which is a straight line.

Problem 2:

Determine all values of \alpha for which the point (\alpha,\alpha^{2}) lies inside the triangle formed by the lines 2x+3y-1=0, x+2y-3=0, 5x-6y-1=0.

Solution 2:

Solving equations of the lines two at a time, we get the vertices of the given triangle as: A(-7,5), B(1/3,1/9) and C(5/4, 7/8).

So, AB is the line 2x+3y-1=0, AC is the line x+2y-3=0 and BC is the line 5x-6y-1=0

Let P(\alpha,\alpha^{2}) be a point inside the triangle ABC. (please do draw it on a sheet of paper, if u want to understand this solution further.) Since A and P lie on the same side of the line 5x-6y-1=0, both 5(-7)-6(5)-1 and 5\alpha-6\alpha^{2}-1 must have the same sign.

\Longrightarrow 5\alpha-6\alpha^{2}-1<0 or 6\alpha^{2}-5\alpha+1>0 which in turn \Longrightarrow (3\alpha-1)(2\alpha-1)>0 which in turn \Longrightarrow either \alpha<1/3 or \alpha>1/2….call this relation I.

Again, since B and P lie on the same side of the line x+2y-3=0, (1/3)+(2/9)-3 and \alpha+2\alpha^{2}-3 have the same sign.

\Longrightarrow 2\alpha^{2}+\alpha-3<0 and \Longrightarrow (2\alpha+3)(\alpha-1)<0, that is, -3/2 <\alpha <1…call this relation II.

Lastly, since C and P lie on the same side of the line 2x+3y-1=0, we have 2 \times (5/4) + 3 \times (7/8) -1 and 2\alpha+3\alpha^{2}-1 have the same sign.

\Longrightarrow 3\alpha^{2}+2\alpha-1>0 that is (3\alpha-1)(\alpha+1)>0

\alpha<-1 or \alpha>1/3….call this relation III.

Now, relations I, II and III hold simultaneously if -3/2 < \alpha <-1 or 1/2<\alpha<1.

Problem 3:

A variable straight line of slope 4 intersects the hyperbola xy=1 at two points. Find the locus of the point which divides the line segment between these two points in the ratio 1:2.

Solution 3:

Let equation of the line be y=4x+c where c is a parameter. It intersects the hyperbola xy=1 at two points, for which x(4x+c)=1, that is, \Longrightarrow 4x^{2}+cx-1=0.

Let x_{1} and x_{2} be the roots of the equation. Then, x_{1}+x_{2}=-c/4 and x_{1}x_{2}=-1/4. If A and B are the points of intersection of the line and the hyperbola, then the coordinates of A are (x_{1}, \frac{1}{x_{1}}) and that of B are (x_{2}, \frac{1}{x_{2}}).

Let R(h,k) be the point which divides AB in the ratio 1:2, then h=\frac{2x_{1}+x_{2}}{3} and k=\frac{\frac{2}{x_{1}}+\frac{1}{x_{2}}}{3}=\frac{2x_{2}+x_{1}}{3x_{1}x_{2}}, that is, \Longrightarrow 2x_{1}+x_{2}=3h…call this equation I.

and x_{1}+2x_{2}=3(-\frac{1}{4})k=(-\frac{3}{4})k….call this equation II.

Adding I and II, we get 3(x_{1}+x_{2})=3(h-\frac{k}{4}), that is,

3(-\frac{c}{4})=3(h-\frac{k}{4}) \Longrightarrow (h-\frac{k}{4})=-\frac{c}{4}….call this equation III.

Subtracting II from I, we get x_{1}-x_{2}=3(h+\frac{k}{4})

\Longrightarrow (x_{1}-x_{2})^{2}=9(h+\frac{k}{4})^{2}

\Longrightarrow \frac{c^{2}}{16} + 1= 9(h+\frac{k}{4})^{2}

\Longrightarrow (h-\frac{k}{4})^{2}+1=9(h+\frac{k}{4})^{2}

\Longrightarrow h^{2}-\frac{1}{2}hk+\frac{k^{2}}{16}+1=9(h^{2}+\frac{1}{2}hk+\frac{k^{2}}{16})

\Longrightarrow 16h^{2}+10hk+k^{2}-2=0

so that the locus of R(h,k) is 16x^{2}+10xy+y^{2}-2=0

More later,

Nalin Pithwa.

By nkpithwa | Also posted in IITJEE Mains, ISI Kolkatta Entrance Exam, KVPY | Comments (0)

Cartesian system and straight lines: IITJEE Mains: Problem solving skills

September 22, 2017 – 3:06 am

Problem 1:

The line joining A(b\cos{\alpha},b\sin{\alpha}) and B(a\cos{\beta},a\sin{\beta}) is produced to the point M(x,y) so that AM:MB=b:a, then find the value of x\cos{\frac{\alpha+\beta}{2}}+y\sin{\frac{\alpha+\beta}{2}}.

Solution 1:

As M divides AB externally in the ratio b:a, we have x=\frac{b(a\cos{\beta})-a(b\cos{\alpha})}{b-a} and y=\frac{b(a\sin{\beta})-a(b\sin{\alpha})}{b-a} which in turn

\Longrightarrow \frac{x}{y} = \frac{\cos{\beta}-cos{\alpha}}{\sin{\beta}-\sin{\alpha}}

= \frac{2\sin{\frac{\alpha+\beta}{2}}\sin{\frac{\alpha-\beta}{2}}}{2\cos{\frac{\alpha+\beta}{2}}\sin{\frac{\beta-\alpha}{2}}}

\Longrightarrow x\cos{\frac{\alpha+\beta}{2}}+y\sin{\frac{\alpha+\beta}{2}}=0

Problem 2:

If the circumcentre of a triangle lies at the origin and the centroid in the middle point of the line joining the points (a^{2}+1,a^{2}+1) and (2a,-2a), then where does the orthocentre lie?

Solution 2:

From plane geometry, we know that the circumcentre, centroid and orthocentre of a triangle lie on a line. So, the orthocentre of the triangle lies on the line joining the circumcentre (0,0) and the centroid (\frac{(a+1)^{2}}{2},\frac{(a-1)^{2}}{2}), that is, y.\frac{(a+1)^{2}}{2} = x.\frac{(a-1)^{2}}{2}, or (a-1)^{2}x-(a+1)^{2}y=0. That is, the orthocentre lies on this line.

Problem 3:

If a, b, c are unequal and different from 1 such that the points (\frac{a^{3}}{a-1},\frac{a^{2}-3}{a-1}), (\frac{b^{3}}{b-1},\frac{b^{2}-3}{b-1}) and (\frac{c^{3}}{c-1},\frac{c^{2}-3}{c-1}) are collinear, then which of the following option is true?

a: bc+ca+ab+abc=0

b: a+b+c=abc

c: bc+ca+ab=abc

d: bc+ca+ab-abc=3(a+b+c)

Solution 3:

Suppose the given points lie on the line lx+my+n=0 then a, b, c are the roots of the equation :

lt^{3}+m(t^{2}-3)+n(t-1)=0, or

lt^{3}+mt^{2}+nt-(3m+n)=0

\Longrightarrow a+b+c=-\frac{m}{l} and ab+bc+ca=\frac{n}{l}, that is, abc=(3m+n)/l

Eliminating l, m, n, we get abc=-3(a+b+c)+bc+ca+ab

\Longrightarrow bc+ca+ab-abc=3(a+b+c), that is, option (d) is the answer.

Problem 4:

If p, x_{1}, x_{2}, \ldots, x_{i}, \ldots and q, y_{1}, y_{2}, \ldots, y_{i}, \ldots are in A.P., with common difference a and b respectively, then on which line does the centre of mean position of the points A_{i}(x_{i},y_{i}) with i=1,2,3 \ldots, n lie?

Solution 4:

Note: Centre of Mean Position is (\frac{\sum{xi}}{n},\frac{\sum {yi}}{n}).

Let the coordinates of the centre of mean position of the points A_{i}, i=1,2,3, \ldots,n be (x,y) then

x=\frac{x_{1}+x_{2}+x_{3}+\ldots + x_{n}}{n} and y=\frac{y_{1}+y_{2}+\ldots + y_{n}}{n}

\Longrightarrow x = \frac{np+a(1+2+\ldots+n)}{n}, y=\frac{nq+b(1+2+\ldots+n)}{n}

\Longrightarrow x=p+ \frac{n(n+1)}{2n}a and y=q+ \frac{n(n+1)}{2n}b

\Longrightarrow x=p+\frac{n+1}{2}a, and y=q+\frac{n+1}{2}b

\Longrightarrow 2\frac{(x-p)}{a}=2\frac{(y-q)}{b} \Longrightarrow bx-ay=bp-aq, that is, the CM lies on this line.

Problem 5:

The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If p and q are the intercepts of the line L on the new axes, then what is the value of \frac{1}{a^{2}} - \frac{1}{p^{2}} + \frac{1}{b^{2}} - \frac{1}{q^{2}}?

Solution 5:

Equation of the line L in the two coordinate systems is \frac{x}{a} + \frac{y}{b}=1, and \frac{X}{p} + \frac{Y}{q}=1 where (X,Y) are the new coordinate of a point (x,y) when the axes are rotated through a fixed angle, keeping the origin fixed. As the length of the perpendicular from the origin has not changed.

\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}=\frac{1}{\sqrt{\frac{1}{p^{2}} + \frac{1}{q^{2}}}}

\Longrightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{p^{2}} + \frac{1}{q^{2}}

or \frac{1}{a^{2}} - \frac{1}{p^{2}} + \frac{1}{b^{2}} - \frac{1}{q^{2}}=0. So, the value is zero.

Problem 6:

Let O be the origin, A(1,0) and B(0,1) and P(x,y) are points such that xy>0 and x+y<1, then which of the following options is true:

a: P lies either inside the triangle OAB or in the third quadrant

b: P cannot lie inside the triangle OAB

c: P lies inside the triangle OAB

d: P lies in the first quadrant only.

Solution 6:

Since xy>0, P either lies in the first quadrant or in the third quadrant. The inequality x+y<1 represents all points below the line x+y=1. So that xy>0 and x+y<1 imply that either P lies inside the triangle OAB or in the third quadrant.

Problem 7:

An equation of a line through the point (1,2) whose distance from the point A(3,1) has the greatest value is :

option i: y=2x

option ii: y=x+1

option iii: x+2y=5

option iv: y=3x-1

Solution 7:

Let the equation of the line through (1,2) be y-2=m(x-1). If p denotes the length of the perpendicular from (3,1) on this line, then p=|\frac{2m+1}{\sqrt{m^{2}+1}}|

\Longrightarrow p^{2}=\sqrt{\frac{4m^{2}+4m+1}{m^{2}+1}}=4+ \frac{4m-3}{m^{2}+1}=s, say

then p^{2} is greatest if and only if s is greatest.

Now, \frac{ds}{dm} = \frac{(m^{2}+1)(4)-2m(4m-3)}{(m^{2}+1)^{2}} = \frac{-2(2m-1)(m-2)}{(m^{2}+1)^{2}}

\frac{ds}{dm} = 0 so that \Longrightarrow m = \frac{1}{2}, 2. Also, \frac{ds}{dm}<0, if m<\frac{1}{2}, and

\frac{ds}{dm} >0, if 1/2<m<2

and \frac{ds}{dm} <0, if m>2. So s is greatest for m=2. And, thus, the equation of the required line is y=2x.

Problem 8:

The points A(-4,-1), B(-2,-4), Slatex C(4,0)$ and D(2,3) are the vertices of a :

option a: parallelogram

option b: rectangle

option c: rhombus

option d: square.

Note: more than one option may be right. Please mark all that are right.

Solution 8:

Mid-point of AC = (\frac{-4+4}{2},\frac{-1+0}{2})=(0, \frac{-1}{2})

Mid-point of BD = (\frac{-2+2}{2},\frac{-4+3}{2})=(0,\frac{-1}{2})

\Longrightarrow the diagonals AC and BD bisect each other.

\Longrightarrow ABCD is a parallelogram.

Next, AC= \sqrt{(-4-4)^{2}+(-1+0)^{2}}=\sqrt{64+1}=\sqrt{65} and BD=\sqrt{(-2-2)^{2}+(-4+3)^{2}}=\sqrt{16+49}=\sqrt{65} and since the diagonals are also equal, it is a rectangle.

As AB=\sqrt{(-4+2)^{2}+(-1+4)^{2}}=\sqrt{13} and BC=\sqrt{(-2-4)^{2}+(-4)^{2}}=\sqrt{36+16}=sqrt{52}, the adjacent sides are not equal and hence, it is neither a rhombus nor a square.

Problem 9:

Equations (b-c)x+(c-a)y+(a-b)=0 and (b^{3}-c^{3})x+(c^{3}-a^{3})y+a^{3}-b^{3}=0 will represent the same line if

option i: b=c

option ii: c=a

option iii: a=b

option iv: a+b+c=0

Solution 9:

The two lines will be identical if there exists some real number k, such that

b^{3}-c^{3}=k(b-c), and c^{3}-a^{3}=k(c-a), and a^{3}-b^{3}=k(a-b).

\Longrightarrow b-c=0 or b^{2}+c^{2}+bc=k

\Longrightarrow c-a=0 or c^{2}+a^{2}+ac=k, and

\Longrightarrow a-b=0 or a^{2}+b^{2}+ab=k

That is, b=c or c=a, or a=b.

Next, b^{2}+c^{2}+bc=c^{2}+a^{2}+ca \Longrightarrow b^{2}-a^{2}=c(a-b). Hence, a=b, or a+b+c=0.

Problem 10:

The circumcentre of a triangle with vertices A(a,a\tan{\alpha}), B(b, b\tan{\beta}) and C(c, c\tan{\gamma}) lies at the origin, where 0<\alpha, \beta, \gamma < \frac{\pi}{2} and \alpha + \beta + \gamma = \pi. Show that it’s orthocentre lies on the line 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}x-4\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}y=y

Solution 10:

As the circumcentre of the triangle is at the origin O, we have OA=OB=OC=r, where r is the radius of the circumcircle.

Hence, OA^{2}=r^{2} \Longrightarrow a^{2}+a^{2}\tan^{2}{\alpha}=r^{2} \Longrightarrow a = r\cos{\alpha}

Therefore, the coordinates of A are (r\cos{\alpha},r\sin{\alpha}). Similarly, the coordinates of B are (r\cos{\beta},r\sin{\beta}) and those of C are (r\cos{\gamma},r\sin{\gamma}). Thus, the coordinates of the centroid G of \triangle ABC are

(\frac{1}{3}r(\cos{\alpha}+\cos{\beta}+\cos{\gamma}),\frac{1}{3}r(\sin{\alpha}+\sin{\beta}+\sin{\gamma})).

Now, if P(h,k) is the orthocentre of \triangle ABC, then from geometry, the circumcentre, centroid, and the orthocentre of a triangle lie on a line, and the slope of OG equals the slope of OP.

Hence, \frac{\sin{\alpha}+\sin{\beta}+\sin{\gamma}}{\cos{\alpha}+\cos{\beta}+\cos{\gamma}}=\frac{k}{h}

\Longrightarrow \frac{4\cos{(\frac{\alpha}{2})}\cos{(\frac{\beta}{2})}\cos{(\frac{\gamma}{2})}}{1+4\sin{(\frac{\alpha}{2})}\sin{(\frac{\beta}{2})}\sin{(\frac{\gamma}{2})}}= \frac{k}{h}

because \alpha+\beta+\gamma=\pi.

Hence, the orthocentre P(h,k) lies on the line

4\cos{(\frac{\alpha}{2})}\cos{(\frac{\beta}{2})}\cos{(\frac{\gamma}{2})}x-4\sin{(\frac{}{})}\sin{(\frac{\beta}{2})}\sin{(\frac{\gamma}{2})}y=y.

Hope this gives an assorted flavour. More stuff later,

Nalin Pithwa.

By nkpithwa | Also posted in IITJEE Mains, ISI Kolkatta Entrance Exam, KVPY | Comments (0)

Cartesian System of Rectangular Co-ordinates and Straight Lines: Basics for IITJEE Mains

September 19, 2017 – 1:56 pm

I. Results regarding points in a plane:

1a) Distance Formula:

The distance between two points P(x_{1},y_{1}) and Q(x_{2},y_{2}) is given by PQ=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}. The distance from the origin O(0,0) to the point P(x_{1},y_{1}) is OP=\sqrt{x_{1}^{2}+y_{1}^{2}}.

1b) Section Formula:

If R(x,y) divides the join of P(x_{1},y_{1}) and Q(x_{2},y_{2}) in the ratio m:n with m>0, n>0, m \neq n, then

x = \frac{mx_{2} \pm nx_{1}}{m \pm n}, and y = \frac{my_{2} \pm ny_{1}}{m \pm n}

The positive sign is taken for internal division and the negative sign for external division. The mid-point of P(x_{1},y_{1}) and Q(x_{2},y_{2}) is (\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}) which corresponds to internal division, when m=n. Note that for external division m \neq n.

1c) Centroid of a triangle:

If G(x,y) is the centroid of the triangle with vertices A(x_{1},y_{1}), B(x_{2},y_{2}) and C(x_{3},y_{3}) then x=\frac{x_{1}+x_{2}+x_{3}}{3} and y=\frac{y_{1}+y_{2}+y_{3}}{3}

1d) Incentre of a triangle:

If I(x,y) is the incentre of the triangle with vertices A(x_{1},y_{1}), B(x_{2},y_{2}), and C(x_{3},y_{3}), then

x=\frac{ax_{1}+bx_{2}+cx_{3}}{a+b+c}, y=\frac{ay_{1}+by_{2}+cy_{3}}{a+b+c}, a, b and c being the lengths of the sides BC, CA and AB, respectively of the triangle ABC.

1e) Area of triangle:

ABC with vertices A(x_{1},y_{1}), B(x_{2},y_{2}), and C(x_{3},y_{3}) is \frac{1}{2}\left|\begin{array}{ccc} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})],

and is generally denoted by \triangle. Note that if one of the vertex (x_{3},y_{3}) is at O(0,0), then \triangle = \frac{1}{2}|x_{1}y_{2}-x_{2}y_{1}|.

Note: When A, B, and C are taken as vertices of a triangle, it is assumed that they are not collinear.

1f) Condition of collinearity:

Three points A(x_{1},y_{1}), B(x_{2},y_{2}), and C(x_{3},y_{3}) are collinear if and only if

\left | \begin{array}{ccc} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} \right |=0

1g) Slope of a line:

Let A(x_{1},y_{1}) and B(x_{2},y_{2}) with x_{1} \neq x_{2} be any two points. Then, the slope of the line joining A and B is defined as

m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \tan{\theta}

where \theta is the angle which the line makes with the positive direction of the x-axis, 0 \deg \leq \theta \leq 180 \deg, except at \theta=90 \deg. Which is possible only if x_{1}=x_{2} and the line is parallel to the y-axis.

1h) Condition for the points Z_{k}=x_{k}+iy_{k}, (k=1,2,3) to form an equilateral triangle is

Z_{1}^{2}+Z_{2}^{2}+Z_{3}^{2}=Z_{1}Z_{2}+Z_{2}Z_{3}+Z_{3}Z_{1}

II) Standard Forms of the Equation of a Line:

  1. An equation of a line parallel to the x-axis is y=k and that of the x-axis itself is y=0.
  2. An equation of a line parallel to the y-axis is x=h and that of the y-axis itself is x=0.
  3. An equation of a line passing through the origin and (a) making an angle \theta with the positive direction of the x-axis is y=x\tan{\theta}, and (b) having a slope m is y=mx, and (c) passing through the point x_{1}y=y_{1}x.
  4. Slope-intercept form: An equation of a line with slope m and making an intercept c on the y-axis is y=mx+c.
  5. Point-slope form: An equation of a line with slope m and passing through (x_{1},y_{1}) is y-y_{1}=x-x_{1}.
  6. Two-point form: An equation of a line passing through the points (x_{1},y_{1}) and (x_{2},y_{2}) is \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{x-x_{1}}{x_{2}-x_{1}}.
  7. Intercept form: An equation of a line making intercepts a and b on the x-axis and y-axis respectively, is \frac{x}{a} + \frac{y}{b}=1.
  8. Parametric form: An equation of a line passing through a fixed point A(x_{1},y_{1}) and making an angle \theta with 0 \leq \theta \leq \pi with \theta \neq \pi/2 with the positive direction of the x-axis is \frac{x-x_{1}}{\cos{\theta}}= \frac{y-y_{1}}{\sin{\theta}} = r where r is the distance of any point P(x,y) on the line from the point A(x_{1},y_{1}). Note that x=x_{1}+r\cos{\theta} and y=y_{1}+r\sin{\theta}.
  9. Normal form: An equation of a line such that the length of the perpendicular from the origin on it is p and the angle which this perpendicular makes with the positive direction of the x-axis is \alpha, is x\cos{\alpha}+y\sin{\alpha}=p.
  10. General form: In general, an equation of a straight line is of the form ax+by+c=0, where a, b, and c are real numbers and a and b cannot both be zero simultaneously. From this general form of the equation of the line, we can calculate the following: (i) the slope is -\frac{a}{b} (ii) the intercept on the x-axis is -\frac{c}{a} with a \neq 0 and the intercept on the y-axis is -\frac{c}{b} with b \neq 0 (iii) p=\frac{|c|}{\sqrt{a^{2}+b^{2}}} and \cos{\alpha} = \pm \frac{|a|}{\sqrt{a^{2}+b^{2}}} and \sin{\alpha}=\pm \frac{|b|}{\sqrt{a^{2}+b^{2}}}, the positive sign being taken if c is negative and vice-versa (iv) If p_{1} denotes the length of the perpendicular from (x_{1},y_{1}) on this line, then p_{1}=\frac{|ax_{1}+by_{1}+c|}{|\sqrt{a^{2}+b^{2}}|} and (v) the points (x_{1},y_{1}) and (x_{2},y_{2}) lie on the same side of the line if the expressions ax_{1}+by_{1}+c and ax_{2}+by_{2}+c have the same sign, and on the opposite side if they have the opposite signs.

III) Some results for two or more lines:

  1. Two lines given by the equations ax+by+c=0 and a^{'}x+b^{'}y+c^{'}=0 are
    • parallel (that is, their slopes are equal) if ab^{'}=a^{'}b
    • perpendicular (that is, the product of their slopes is -1) if aa^{'}+bb^{'}=0
    • identical if ab^{'}c^{'}=a^{'}b^{'}c=a^{'}c^{'}b
    • not parallel, then
      • angle \theta between them at their point of intersection is given by \tan{\theta}= \pm \frac{m-m^{'}}{1+mm^{'}} = \pm \frac{a^{'}b-ab^{'}}{aa^{'}+bb^{'}} where m, m^{'} being the slopes of the two lines.
      • the coordinates of their points of intersection are (\frac{bc^{'}-c^{'}b}{ab^{'}-a^{'}b}, \frac{ca^{'}-c^{'}a}{ab^{'}-a^{'}b})
      • An equation of any line through their point of intersection is (ax+by+c) + \lambda (a^{'}x+b^{'}y+c^{'})=0 where \lambda is a real number.
  2. An equation of a line parallel to the line ax+by+c=0 is ax+by+c^{'}=0, and the distance between these lines is \frac{|c-c^{'}|}{\sqrt{a^{2}+b^{2}}}
  3. The three lines a_{1}x+b_{1}y+c_{1}=0, a_{2}x+b_{2}y+c_{2}=0 and a_{3}x+b_{3}y+c_{3}=0 are concurrent (intersect at a point) if and only if \left | \begin{array}{ccc} a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array} \right|=0
  4. Equations of the bisectors of the angles between two intersecting lines ax+by+c=0 and a^{'}x+b^{'}y+c^{'}=0 are \frac{ax+by+c}{\sqrt{a^{2}+b^{2}}}=\pm \frac{a^{'}x+b^{'}y+c^{'}}{\sqrt{a^{'2}+b^{'2}}}. Any point on the bisectors is equidistant from the given lines. If \phi is the angle between one of the bisectors and one of the lines ax+by+c=0 such that |\tan{\phi}|<1, that is, -\frac{\pi}{4} < \phi < \frac{\pi}{4}, then that bisector bisects the acute angle between the two lines, that is, it is the acute angle bisector of the two lines. The other equation then represents the obtuse angle bisector between the two lines.
  5. Equations of the lines through (x_{1},y_{1}) and making an angle \phi with the line ax+by+c=0, b \neq 0 are y-y_{1}=m_{1}(x-x_{1}) where m_{1}=\frac{\tan{\theta}-\tan{\phi}}{1+\tan{\theta}\tan{\phi}} and y-y_{1}=m_{2}(x-x_{1}) where m_{2}=\frac{\tan{\theta}+\tan{\phi}}{1-\tan{\theta}\tan{\phi}} where \tan{\theta}=-\frac{a}{b} is the slope of the given line. Note that m_{1}=\tan{(\theta-\phi)} and m_{2}=\tan{(\theta + \phi)} and when b=0, \theta=\frac{\pi}{2}.

IV) Some Useful Points:

To show that A, B, C, D are the vertices of a

  1. parallelogram: show that the diagonals AC and BD bisect each other.
  2. rhombus: show that the diagonals AC and BD bisect each other and a pair of adjacent sides, say, AB and BC are equal.
  3. square: show that the diagonals AC and BD are equal and bisect each other, a pair of adjacent sides, say AB and BC are equal.
  4. rectangle: show that the diagonals AC and BD are equal and bisect each other.

V) Locus of a point:

To obtain the equation of a set of points satisfying some given condition(s) called locus, proceed as follows:

  • Let P(h,k) be any point on the locus.
  • Write the given condition involving h and k and simplify. If possible, draw a figure.
  • Eliminate the unknowns, if any.
  • Replace h by x and k by y and obtain an equation in terms of (x,y) and the known quantities. This is the required locus.

VI) Change of Axes:

  1. Rotation of Axes: if the axes are rotated through an angle \theta in the anti-clockwise direction keeping the origin fixed, then the coordinates (X,Y) of a point P(x,y) with respect to the new system of coordinates are given by X=x\cos{\theta}+y\sin{\theta} and Y=y\cos{\theta}-x\sin{\theta}.
  2. Translation of Axes: the shifting of origin of axes without rotation of axes is called translation of axes. If the origin (0,0) is shifted to the point (h,k) without rotation of the axes then the coordinates (X,Y) of a point P(x,y) with respect to the new system of coordinates are given by X=x-h and Y=y-k.

I hope to present some solved sample problems with solutions soon.

Nalin Pithwa.

By nkpithwa | Also posted in IITJEE Mains, KVPY | Comments (0)

Conics: Homework for IITJEE Mains: Hausaufgabe Grundlagen konischen!

September 16, 2017 – 5:15 am

Aber das ist English ! 🙂

Question 1:

Find the equation to that tangent to the parabola y^{2}=7x which is parallel to the straight line 4y-x+3=0. Find also its point of contact.

Question 2:

If P, Q and R are three points on the parabola y^{2}=4ax whose coordinates are in geometric progression, prove that the tangents at P and R meet on the ordinate of Q.

Question 3:

PNP^{'} is a double ordinate of the parabola y^{2}=4ax. Prove that the locus of the point of intersection of the normal at P and straight line through P^{'} parallel to the axis is the parabola y^{2}=4a(x-4a).

Question 4:

Show that in a parabola, the length of the focal chord varies inversely as the square of the distance of the vertex of the parabola from the focal chord.

Question 5:

Prove that the equation y^{2}+2ax+2by+c=0 represents a parabola whose axis is parallel to the x-axis.. Find its vertex.

Question 6:

If the line y=3x+1 touches the parabola y^{2}=4ax, find the length of the latus rectum.

Question 7a:

Prove that the circle described on any focal chord of a parabola as the diameter touches the directrix of the parabola.

Question 7b:

Show that the locus of the point, such that two of the normals drawn from it to the parabola y^{2}=4ax coincide is 27ay^{2}=4(x-2a)^{3}.

Question 7c:

If the normals at three points A, B and C on the parabola y^{2}=4ax pass through the point S(h,k) and cut the axis of the parabola in P, Q and R so that OP, OQ, OR are in AP, O being the vertex of the parabola, prove that the locus of the point S is 27ay^{}2=2(x-2a)^{2}.

Question 8:

If P, Q and R be three conormal points on the parabola y^{2}=4ax, the normals at which pass through the point T, and S is the focus of the parabola, then prove that SP.SQ.SR=a.ST^{2}

Question 9:

The tangents at P and Q to the parabola y^{2}=4ax meet in T and the corresponding normals meet in R. If the locus of T is a straight line parallel to the axis of the parabola, prove that the locus of R is a straight line normal to the parabola.

Question 10a:

A variable chord PQ of the parabola y^{2}=4x is drawn parallel to the line y=x. If the parameters of the points P and Q on the parabola be t_{1} and t_{2}, then t_{1}+t_{2}=2. Also, show that the locus of the point of intersection of the normals at P and Q is 2x-y=12, which is itself a normal to the parabola.

Question 10b:

PQ is a chord of the parabola y^{2}=36x whose right bisector meets the axis in M and the ordinate of the mid-point of PQ meets the axis in L. Show that LM is constant and find LM.

Question 11:

Prove that the locus of a point P such that the slopes m_{1}, m_{2}, m_{3} of the three normals drawn to the parabola y^{2}=4x from P be connected by the relation \arctan{m_{1}^{2}}+\arctan{m_{2}^{2}} + \arctan{m_{3}^{2}=\alpha} is x^{2}\tan{\alpha}-y^{2}+2(1-2\tan{\alpha})x+(3\tan{\alpha}-4)=0.

Question 12a:

If the perpendicular drawn from P on the polar of P with respect to the parabola, y^{2}=4by, prove that the locus of P is the straight line 2ax+by+4a^{2}=0.

Question 12b:

Prove that the locus of the poles of the tangent to the parabola y^{2}=4ax w.r.t. the circle x^{2}+y^{2}=2ax is x^{2}+y^{2}=ax.

Question 13:

Tangents are drawn to the parabola y^{2}=4ax at the points P and Q whose inclination to the axis are \theta_{1}, \theta_{2}. If A be the vertex of the parabola and the circles on AP and AQ as diameters intersect in R and AR be inclined at an angle \phi to the axis, then prove that \cot{\theta_{1}}+\cot_{\theta_{2}}+2\tan_{\phi}=0.

Question 14:

Through the vertex O of a parabola y^{2}=4x chords OP and OQ are drawn at right angles to one another. Prove that for all positions of P, PQ cuts the axis of the parabola at a fixed point. Also, find the locus of the middle point of PQ.

Question 15a:

Prove that the area of a triangle inscribed in a parabola is twice the area of the triangle formed by the tangents at the vertices of the triangle.

Question 15b:

Normals are drawn to the parabola y^{2}=4ax at points A, B, and C whose parameter are t_{1}, t_{2}, t_{3} respectively. If these normals enclose a triangle PQR, then prove that its area is \frac{a^{2}}{2}(t_{1}-t_{2})(t_{2}-t_{3})(t_{3}-t_{1})(t_{1}+t_{2}+t_{3})^{2}. Also, prove that \triangle PQR=\triangle ABC (t_{1}+t_{2}+t_{3})^{2}.

Question 16:

Find the locus of the middle points of the chords of the ellipse (\frac{x^{2}}{a^{2}})+(\frac{y^{2}}{b^{2}})=1 which are drawn through the positive end of the minor axis.

Question 17:

Prove that the sum of the squares of the perpendiculars on any tangent to the ellipse (\frac{x^{2}}{a^{2}})+(\frac{y^{2}}{b^{2}})=1 from the points on the minor axis, each at a distance \sqrt{a^{2}-b^{2}} from the centre, is 2a^{2}.

Question 18:

If P(a\cos{\alpha}, b\sin{\alpha}) and Q(a\cos{\beta}, b\sin{\beta}) are two variable points on the ellipse (\frac{x^{2}}{a^{2}})+(\frac{y^{2}}{b^{2}})=1 such that \alpha+\beta=2\gamma (some constant), then prove that the tangent at (a\cos{\gamma},b\sin{\gamma}) is parallel to PQ.

Question 19:

P is a variable point on the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1 whose foci are the points S_{1}, S_{2}, and the eccentricity is e. Prove that the locus of incentre of \triangle PS_{1}S_{2} is an ellipse whose eccentricity is \sqrt{\frac{2e}{1+e}}.

Question 20:

Consider the family of circles x^{2}+y^{2}=r^{2}, 2<r<5. If in the first quadrant, the common tangent to a circle of this family and the ellipse 4x^{2}+25y^{2}=100 meets the coordinate axes at A and B, then find the equation of the locus of the mid-point of AB.

Question 21:

Let P be a point on the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1, 0<b<a. Let the line parallel to y-axis passing through P meet the circle x^{2}+y^{2}=a^{2} at the point Q such that P and Q are on the same side of the x-axis. For two positive real numbers, r and s, find the locus of the point R on PQ such that PR:RQ=r:s as P varies over the ellipse.

Question 22:

If the eccentric angles of points P and Q on the ellipse be \theta and \frac{\pi}{2} + \theta and \alpha be the angle between the normals at P and Q, then prove that the eccentricity e is given by 2\sqrt{1-e^{2}}=e^{2}=e^{2}\sin^{2}{2\theta}\tan{\alpha}.

Question 23:

A series of hyperbolas are such that the length of their transverse axis is 2a. Prove that the locus of a point P on each, such that its distance from transverse axis is equal to its distance from an asymptote is the curve:

(x^{2}-y^{2})^{2}=4x^{2}(x^{2}-a^{2}).

Question 24:

A variable line of slope 4 intersects the hyperbola xy=1 at two points. Find the locus of the point which divides the line segment between these two points in the ratio 1:2.

Question 25:

If the tangent at the point (p,q) on the hyperbola \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}=1 cuts the auxillary circle in points whose coordinates are y_{1} and y_{2}, then show that q is harmonic mean of y_{1} and y_{2}.

Question 26:

Show that the locus of poles with respect to the parabola y^{2}=4ax of the tangents to the hyperbola x^{2}-y^{2}=a^{2} to the ellipse 4x^{2}+y^{2}=4a^{2}.

Question 27:

The point P on the hyperbola with focus S is such that the tangent at P, the latus rectum through S and one asymptote are concurrent. Prove that SP is parallel to other asymptote.

Question 28:

If a triangle is inscribed in a rectangular hyperbola, prove that the orthocentre of the triangle lies on the curve.

Question 29:

A series of chords of the hyperbola \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}=1 touch the circle on the line joining the foci as diameter. Show that the locus of the poles of these chords with respect to the hyperbola is \frac{x^{2}}{a^{4}} - \frac{y^{2}}{b^{4}} = \frac{1}{a^{2}+b^{2}}.

Question 30:

Prove that the chord of the hyperbola which touches the conjugate hyperbola is bisected at the point of contact.

Cheers,

Nalin Pithwa.

By nkpithwa | Also posted in IITJEE Mains | Comments (0)

Conics: Co-ordinate Geometry for IITJEE Mains: Basics 6

September 16, 2017 – 3:00 am

Question 1:

Find the locus of the point of intersection of tangents to the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, which are at right angles.

Solution 1:

Any tangent to the ellipse is y=mx + \sqrt{a^{2}m^{2}+b^{2}} …call this equation I.

Equation of the tangent perpendicular to this tangent is y=-\frac{1}{m}x + \sqrt{\frac{a^{2}}{m^{2}}+b^{2}}…call this Equation II.

The locus of the point of intersection of I and II is obtained by eliminating m between these equations. Squaring and adding we get:

(y-mx)^{2}+(my+x)^{2}=a^{2}m^{2}+b^{2}+a^{2}+b^{2}m^{2}

\Longrightarrow (1+m^{2})(x^{2}+y^{2})=(1+m^{2})(a^{2}+b^{2})

\Longrightarrow x^{2}+y^{2}=a^{2}+b^{2}

which is a circle with its centre at the centre of the ellipse and radius equal to the length of the line joining the ends of the major and minor axis. This circle is called the director circle of the ellipse. 🙂

Question 2:

A tangent to the ellipse x^{2}+4y^{2}=4 meets the ellipse x^{2}+2y^{2}=6 at P and Q. Prove that the tangents at P and Q of the ellipse x^{2}+2y^{2}=6 are at right angles.

Solution 2:

Let the tangent at R(2\cos{\theta}, \sin{\theta}) to the ellipse x^{2}+4y^{2}=4 (equation I) meet the ellipse x^{2}+2y^{2}=6 (equation II) at P and Q.

Let the tangents at P and Q to II intersect at the point S(\alpha, \beta). Then, PQ is the chord of contact of the point S(\alpha, \beta) with respect to II and so its equation is \alpha x + 2\beta y =6 ( Equation III).

PQ is also the tangent at R(2\cos{\theta},\sin{\theta}) to I and so its equation can be written as

(2\cos{\theta})x+(4\sin{\theta})y=4 (Equation IV)

Comparing III and IV, we get

\frac{2\cos{\theta}}{\alpha} = \frac{4\sin{\theta}}{2\beta} = \frac{4}{6}

\Longrightarrow \cos{\theta}=\frac{\alpha}{3}, \sin{\theta}=\frac{\beta}{3}

\Longrightarrow \frac{\alpha^{2}}{9} + \frac{\beta^{2}}{9} =1, \Longrightarrow \alpha^{2}+\beta^{2}=9,

the locus of S(\alpha, \beta) is x^{2}+y^{2}=9 or x^{2}+y^{2}=6+3, which is the director circle of the ellipse II. Hence, the tangents at P and Q to the ellipse II are at right angles (using the previous example). 🙂

Question 3:

Let d be the perpendicular distance from the centre of the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1 to the tangent drawn at a point P on the ellipse. If F_{1} and F_{2} are the two foci of the ellipse, then prove that (PF_{1}-PF_{2})^{2}=4a^{2}(1-\frac{b^{2}}{a^{2}}).

Solution 3:

Equation of the tangent at the point P(a\cos{\theta},b\sin{\theta}) on the given ellipse is \frac{x\cos{\theta}}{a} + \frac{y\sin{\theta}}{b}=1. Thus,

d= | \frac{-1}{\sqrt{\frac{\cos^{2}{\theta}}{a^{2}} + \frac{\sin^{2}{\theta}}{b^{2}}}}|

d^{2}=\frac{a^{2}b^{2}}{b^{2}\cos^{2}{\theta}+a^{2}\sin^{2}{\theta}}.

We know that PF_{1}+PF_{2}=2a

\Longrightarrow (PF_{1}-PF_{2})^{2}=(PF_{1}+PF_{2})^{2}-4PF_{1}PF_{2}…call this equation I.

Also, (PF_{1}PF_{2})^{2}=[(a\cos{\theta}-ae)^{2}+(b\sin{\theta})^{2}][(a\cos{\theta}+ae)^{2}+(b\sin{\theta})^{2}],

which in turn equals

[a^{2}(\cos{\theta}-e)^{2}+a^{2}(1-e^{2})\sin^{2}(\theta)][a^{2}(\cos{\theta}+e)^{2}+a^{2}(1-e^{2})\sin^{2}{\theta}],

which in turn equals

a^{4}[(\cos^{2}{\theta}+e^{2})-2e\cos{\theta}+\sin^{2}{\theta}-e^{2}\sin^{2}{\theta}][(\cos^{2}{\theta}+e^{2})+2e\cos{\theta}+\sin^{2}{\theta}-e^{2}\sin^{2}{\theta}],

which in turn equals

a^{4}[1-2e\cos{\theta}+e^{2}\cos^{2}{\theta}][1+2e\cos{\theta}+e^{2}\cos^{2}{\theta}],

which in turn equals

a^{4}[(1+e^{2}\cos^{2}{\theta})^{2}-4e^{2}\cos^{2}{\theta}] = a^{4}[(1-e^{2}\cos^{2}{\theta})^{2}] \Longrightarrow PF_{1}.PF_{2}=a^{2}(1-e^{2}\cos^{2}{\theta})

Now, from I, we get (PF_{1}-PF_{2})^{2}=4a^{2}-4a^{2}(1-e^{2}\cos^{2}{\theta})=4a^{2}e^{2}\cos^{2}{\theta}

Also, 1-\frac{b^{2}}{d^{2}}=1-\frac{b^{2}\cos^{2}{\theta}+a^{2}\sin^{2}{\theta}}{a^{2}}

which in turn equals

\frac{(a^{2}-b^{2})\cos^{2}{\theta}}{a^{2}} = e^{2}\cos^{2}{\theta}.

Hence, (PF_{1}-PF_{2})^{2}=4a^{2}(1-\frac{b^{2}}{d^{2}}). 🙂

More later,

Nalin Pithwa.

By nkpithwa | Also posted in IITJEE Mains | Comments (0)

Conics: Co-ordinate Geometry for IITJEE Mains: Basics 5

September 13, 2017 – 9:11 pm

Question I:

Given the base of a triangle and the ratio of the tangents of half the base angles, prove that the vertex moves on a hyperbola, whose foci are the extremities of the base.

Solution I:

Let the base of the triangle ABC be BC=a, and \frac{\tan{(B/2)}}{\tan{C/2}}=K (constant)

Then, we get \Longrightarrow \sqrt{\frac{(s-c)(s-a)}{s(s-b)}} \times \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}=K

\Longrightarrow \frac{(s-c)}{(s-b)}=K \Longrightarrow \frac{(a+b-c)}{(a-b+c)}=K

$\Longrightarrow latex a+(b-c)=K(a-(b-c))$

2K(b-c)=(K-1)a \Longrightarrow b-c=\frac{(K-1)a}{2K}

Since a and K are given constants, (K-1)a/2K is a constant, say \alpha. Therefore, we get b-c=\alpha, that is, AB-AC=\alpha. We find that the vertex A moves in such a manner that the difference of its distances from two fixed points B and C, which are the extremities of the base is constant. Hence, it describes a hyperbola whose foci are the extremities of the base. For instance, you will get further insight if you try the following: (in previous blog, basics 4, of coordinate geometry) : Show that the difference of focal distances of any point on a hyperbola is constant and equal to the length of the transverse axis.

Question 2:

Show that the angle between the tangents to the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1 and the circle x^{2}+y^{2}=ab at their points of intersection is \arctan{\frac{(a-b)}{\sqrt{ab}}}.

Solution 2:

For the points of intersection, we have \frac{ab-y^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1

\Longrightarrow y^{2}[\frac{1}{b^{2}} - \frac{1}{a^{2}}]=1-\frac{b}{a}

\Longrightarrow y^{2}=\frac{a^{2}b^{2}}{a^{2}-b^{2}}\times \frac{a-b}{a}=\frac{ab^{2}}{a+b}

y= \pm b \sqrt{\frac{a}{a+b}} \Longrightarrow \pm a \sqrt{\frac{b}{a+b}}

Consider the point P(\frac{a\sqrt{b}}{\sqrt{a+b}}, b\frac{\sqrt{a}}{a+b}), intersection of the given ellipse and the circle. Equation of the tangent at P to the circle is \frac{xa\sqrt{b}}{\sqrt{a+b}} + \frac{yb\sqrt{a}}{\sqrt{a+b}}=ab

Slope of this tangent is =-\frac{\sqrt{a}}{\sqrt{b}}. Equation of the tangent at P to the ellipse is \frac{xa\sqrt{b}}{a^{2}\sqrt{a+b}} + \frac{by\sqrt{a}}{b^{2}\sqrt{a+b}}=1 and slope of this tangent is =-\frac{b^{3/2}}{a^{3/2}}. If \alpha be the angle between these tangents, then \tan{\alpha}=\frac{-\frac{b^{3/2}}{a^{3/2}}+ \frac{a^{1/2}}{b^{1/2}}}{1+ \frac{b^{3/2}.a^{1/2}}{a^{3/2}}.b^{-1/2}}=\frac{(a^{2}-b^{2})}{a^{1/2}.b^{1/2}(a+b)}=\frac{(a-b)}{\sqrt{ab}}.

Hence, \alpha = \arctan{\frac{(a-b)}{\sqrt{ab}}}. Note: The angle will be the same at each point of intersection.

Question 3:

A straight line is drawn parallel to the conjugate axis of the hyperbola \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 to meet it and the conjugate hyperbola in the points P and Q respectively. Show that the normals at P and Q to the curves meet on the x-axis.

Solution 3:

Conjugate Hyperbola of the given hyperbola is the hyperbola whose transverse and conjugate axis are respectively the conjugate and transverse axis of the given hyperbola and thus the equation is \frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1.

Now, P(a\sec{\theta},b\tan{\theta}) is a point of the given hyperbola and Q(a\tan{\phi},b\sec{\phi}) is a point on the conjugate hyperbola. Since P and Q lie on the line parallel to the conjugate axis x=0, of the given hyperbola, a\sec{\theta}=a\tan{\phi}, \Longrightarrow \sec{\theta}=\tan{\phi}.

Now, equation of the normal at P to the given hyperbola is y-b\tan{\theta}=-\frac{a\tan{\theta}}{b\sec{\theta}}(x-a\sec{\theta})…call this equation I.

and equation of the normal at Q to the conjugate hyperbola is y-b\sec{\theta}=-\frac{a\sec{\phi}}{b\tan{\phi}}(x-a\tan{\phi})….call this equation II.

Eliminating x from I and II, using \sec {\theta}= \tan{\phi} we get: \frac{b\sec{\theta}}{a\tan{\theta}}(y-b\tan{\theta})=\frac{b\tan{\phi}}{a\sec{\phi}}(y-b\sec{\phi}), which in turn, implies that y=0. Hence, the normals meet on the x-axis.

More later, including homeworks,

Nalin Pithwa.

By nkpithwa | Also posted in IITJEE Mains | Comments (0)

Conics: Co-ordinate Geometry for IITJEE Mains: Problem solving: Basics 4

September 12, 2017 – 10:33 am

Question 1:

If the points of the intersection of the ellipses \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1 and \frac{x^{2}}{\alpha^{2}} + \frac{y^{2}}{\beta^{2}}=1 are the end points of the conjugate diameters of the former, prove that :\frac{a^{2}}{\alpha^{2}} + \frac{b^{2}}{\beta^{2}} = 2

Solution 1:

The locus of the middle points of a system of parallel chords of an ellipse is a line passing through the centre of the ellipse. This is called the diameter of the ellipse and two diameters of the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1 are said to be conjugate if each bisects the chords, parallel to the other. The condition for this is that the product of their slopes should be equal to \frac{-b^{2}}{a^{2}}.

Now, equation of the lines joining the centre (0,0) to the points of intersection of the given ellipses is

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = \frac{x^{2}}{\alpha^{2}} + \frac{y^{2}}{\beta^{2}}

\Longrightarrow (\frac{1}{a^{2}}-\frac{1}{\alpha^{2}})x^{2}+(\frac{1}{b^{2}} - \frac{1}{\beta^{2}})y^{2}=0…call this equation I;

If m_{1}, m_{2} are the slopes of the lines represented by Equation I, then m_{1}m_{2}=\frac{(\frac{1}{a^{2}}-\frac{1}{\alpha^{2}})}{(\frac{1}{b^{2}}-\frac{1}{\beta^{2}})}

Since I represents a pair of conjugate diameters, m_{1}m_{2}=-\frac{b^{2}}{a^{2}}

Thus, a^{2}(\frac{1}{a^{2}}-\frac{1}{\alpha^{2}})+b^{2}(\frac{1}{b^{2}}-\frac{1}{\beta^{2}})=0

\Longrightarrow \frac{a^{2}}{\alpha^{2}} + \frac{b^{2}}{\beta^{2}} =2

Question 2:

Find the locus of the mid-points of the chords of the circle x^{2}+y^{2}=16 which are tangents to the hyperbola, 9x^{2}-16y^{2}=144.

Solution 2:

Let (h,k) be the middle point of a chord of the circle x^{2}+y^{2}=16.

Then, its equation is hx + ky-16=h^{2}+k^{2}-16, that is, hx + ky=h^{2}+k^{2} ….call this equation I.

Let I touch the hyperbola: 9x^{2}-16y^{2}=144

That is, \frac{x^{2}}{16} - \frac{y^{2}}{9}=1 …call this equation II.

at the point (\alpha, \beta) say, then I is identical with

\frac{x\alpha}{16} - \frac{y\beta}{9}=1….call this equation III.

Thus, \frac{\alpha}{16h} = \frac{-\beta}{9k} = \frac{1}{h^{2}+k^{2}}

Since (\alpha, \beta) lies on the hyperbola II,

\frac{1}{16}(\frac{16h}{(h^{2}+k^{2})})^{2}-\frac{1}{9}(\frac{9k}{h^{2}+k^{2}})^{2}=1

\Longrightarrow 16h^{2}-9k^{2}=(h^{2}+k^{2})^{2}.

Hence, the required locus of (h,k) is (x^{2}+y^{2})^{2}=16x^{2}-9y^{2}.

Question 3:

If P be a point on the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1 whose ordinate is y^{'}, prove that the angle between the tangent at P and the focal chord through P is \arctan{(\frac{b^{2}}{aey^{'}})}.

Solution 3:

Let the coordinates of P be (a\cos{\theta}, b\sin{\theta}) so that b\sin{\theta}=y^{'}. Equation of the tangent at P is \frac{x}{a}\cos{\theta}+\frac{y}{b}\sin{\theta}=1.

Slope of the tangent is equal to -\frac{b\cos{\theta}}{a\sin{\theta}}

Slope of the focal chord SP is \frac{b\sin{\theta}-0}{a\cos{\theta}-ae}=\frac{b\sin{\theta}}{a(\cos{\theta})-e}.

If \alpha is the required angle, then \tan{\alpha}=\frac{-\frac{b\cos{\theta}}{a\sin{\theta}}-\frac{b\sin{\theta}}{a(\cos{\theta})-e}}{1-\frac{b\cos{\theta}b\sin{\theta}}{a\sin{\theta}a(\cos{\theta})-e}}

which in turn equals \frac{-b(\cos^{2}{\theta}-e\cos{\theta}+\sin^{2}{\theta})}{a\sin{\theta}(\cos{\theta}-e)} \times \frac{a^{2}(\cos{\theta}-e)}{(a^{2}-b^{2})\cos{\theta}-a^{2}e},

=\frac{-ab(1-e\cos{\theta})}{\sin{\theta}(a^{2}e^{2}\cos{\theta}-a^{2}e)}=\frac{ab(e\cos{\theta}-1)}{a^{2}e\sin{\theta}(e\cos{\theta}-1)}

=\frac{ab}{a^{2}e\sin{\theta}}=\frac{ab.b}{a^{2}e.y^{'}}=\frac{b^{2}}{aey^{'}}

\Longrightarrow \alpha=\arctan{(\frac{b^{2}}{aey^{'}})}

More later,

I hope you like it…my students should be inspired to try Math on their own…initially, it is slow, gradual, painstaking, but the initial “roots” pay very very “rich dividends” later…

-Nalin Pithwa.

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Conics: Co-ordinate Geometry: problem solving for IITJEE Mains: Basics 3

September 11, 2017 – 2:35 pm

One more “nice” problem with solution ! Though, I strongly suggest you try the problem on your own and compare it with the solution given here. You might even come up with a clever approach than the one I try here:

Question:

If r_{1}, r_{2} be the lengths of the radii vectors of the parabola y^{2}=4ax which are drawn at right angles to one another from the vertex. Prove that r_{1}^{4/3}r_{2}^{4/3}=16a^{2}(r_{1}^{2/3}+r_{2}^{2/3}).

Solution:

Let P(r_{1}\cos{\theta}, r_{2}\sin{\theta}) and Q(r_{2}\sin{\theta},r_{2}\cos{\theta}) be two points on the parabola y^{2}=4ax with lengths of the radii vectors r_{1} and r_{2} respectively, then r_{1}^{2}\sin^{2}{\theta}=4ar_{1}\cos{\theta} \Longrightarrow r_{1}=\frac{4a\cos{\theta}}{\sin^{2}{\theta}}

Similarly, r_{2}=\frac{4a\sin{\theta}}{\cos^{2}{\theta}} since radii vectors r_{1} and r_{2} are at right angles.

Hence, r_{1}r_{2}=\frac{16a^{2}}{\sin{\theta}\cos{\theta}}

\Longrightarrow (r_{1}r_{2})^{4/3}=\frac{(16a^{2})^{4/3}}{(\sin{\theta}\cos{\theta})^{4/3}}…call this equation I.

And, r_{1}^{2/3}+r_{2}^{2/3}=(4a)^{2/3}[\frac{\cos^{2/3}{\theta}}{\sin^{4/3}{\theta}} + \frac{\sin^{2/3}{\theta}}{\cos^{4/3}{\theta}}]

which equals \frac{(4a)^{2/3}}{\sin^{4/3}{\theta}\cos^{4/3}{\theta}}[\cos^{2}{\theta}+\sin^{2}{\theta}]….call this equation II.

From I and II, we get : \frac{r_{1}^{4/3}r_{2}^{4/3}}{r_{1}^{2/3}+r_{2}^{2/3}}=\frac{(16a^{2})^{4/3}}{(16a^{2})^{4/3}}=16a^{2}.

More later,

Nalin Pithwa.

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Follow Descartes’ Historically Famous Problems !

May 24, 2017 – 1:59 am

Problem 1:

Three circles touching one another externally have radii r_{1}, r_{2} and r_{3}. Determine the radii of the two circles that can be drawn touching all the three circles.

Problem 2:

Consider a circle, say (numbered 1) of unit radius 1. Inside this circle, two circles are drawn (say, numbered 2 and 3), each of radius \frac{1}{2}, which touch each other externally and the first circle internally. Determine the radius of the fourth circle, which touches circles 2 and 3 externally and circle 1 internally. Determine the radius of the fifth circle, which touches each of the circles 2, 3, and 4 externally. Determine the radius of the sixth circle, which touches circles 2 and 4 externally and circle 1 internally. One might notice that curvature of all such circles drawn within the first circle has integer curvature!

It is such historically famous problems (within scope of IITJEE Mains and IITJEE Advanced Maths) which all students should try to internalize all the concepts of Math for IITJEE. Also, in a similar vein, you should practice deriving all basic formulae, relationships of co-ordinate geometry.

More later,

Nalin Pithwa.

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Some problems on conics (parabola, ellipse, hyperbola) : IITJEE Mains — Basics 2

April 5, 2017 – 5:19 pm

I am solving some “nice” problems below:

Problem 1:

Show that the locus of a point that divides a chord of slope 2 of the parabola y^{2}=4x internally in the ratio 1:2 is a parabola. Find the vertex of this parabola.

Solution 1:

Let P(t_{1}^{2}, 2t_{1}) and Q(t_{2}^{2}, 2t_{2}) be the extremities of the chord with slope 2.

Hence, \frac{2t_{1}-2t_{2}}{t_{1}^{2}-t_{2}^{2}}=2 \Longrightarrow t_{1}+t_{2}=1

Let (t_{1},t_{2}) be co-ordinates of the point which divides PQ in the ratio 1:2. Then,

h=\frac{2t_{1}^{2}+t_{2}^{2}}{3} and k=\frac{4t_{1}+2t_{2}}{3}

\Longrightarrow 3h = 2t_{1}^{2}+(1-t_{1})^{2} and 3k = 4t_{1}+2(1-t_{1})

\Longrightarrow 3h = 3t_{1}^{2} - 2t_{1} + 1 and 3k = 2t_{1}+2

\Longrightarrow 3h = 3(\frac{3k-2}{2})^{2} -2(\frac{4k-2}{2})+1

\Longrightarrow 12h = 3(9k^{2}-12k+4)-12k+8+4

4h = 9k^{2}-16k+8

Hence, the locus of (h,k) is 9y^{2}-16y-4x+8=0

\Longrightarrow (3y-\frac{8}{3})^{2}=4x-8+\frac{64}{9}

\Longrightarrow (y-\frac{8}{9})^{2}=\frac{4}{9}(x-\frac{2}{9}), which is a parabola with vertex (\frac{2}{9}, \frac{8}{9}).

Problem 2:

If P_{1}P_{2} and P_{3}P_{4} are two focal chords of the parabola y^{2}=4ax,  then show that the chords P_{1}P_{3} and P_{2}P_{4} intersect on the directrix of the parabola.

Solution 2:

Let the co-ordinates of P_{i} be (at_{i}^{2},2at_{i}) for i=1,2,3,4.

Since P_{1}P_{2} is a focal chord, t_{1}t_{2}=-1 —– call this Equation I.

Similarly, t_{3}t_{4}=-1 —- call this Equation II.

Equation of P_{1}P_{3} is y(t_{1}+t_{3})=2(x+at_{1}t_{3}) —- call this Equation III.

and that of P_{2}P_{4} is y(t_{2}+t_{4})=2(x+\alpha t_{2}t_{4}) —- call this Equation IV.

Using I and II, IV reduces to y(-\frac{1}{t_{1}}-\frac{1}{t_{2}})=2(x+\frac{a}{t_{1}}t_{3})

that is, -y(t_{1}+t_{3})=2(xt_{1}t_{3}+a) — call this Equation V.

Adding III and V we get:

0=2[x+at_{1}t_{3}+xt_{1}t_{3}+a], which in turn implies, (x+a)(1+t_{1}t_{3})=0, which in turn implies, that x=-a. Hence, III and V intersect on the directrix x+a=0.

More later,

Nalin Pithwa.

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