Category Archives: co-ordinate geometry

Conics: Homework for IITJEE Mains: Hausaufgabe Grundlagen konischen!

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.

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

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.

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

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.

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

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.

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

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.

Follow Descartes’ Historically Famous Problems !

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.

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

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.

Conic Sections (Parabola, Ellipse, Hyperbola): some basic theorems/questions: IITJEE Mains Maths

Problem 1:

(A) Proposition 1: Show  that the tangents at the extremities of a focal chord of a parabola intersect at right angles on the directrix.

Proof 1:

Let t_{1} and t_{2} be the extremities of a focal chord of the parabola y^{2}=4ax. Then, it can be shown that t_{1}t_{2}=-1. (Try this part on your own and let me know; if you can’t produce the proof, I will send it to you.) The equations of the tangents at t_{1} and t_{2} are t_{1}y=x+at_{1}^{2} and t_{2}y=x+at_{2}^{2}. The product of the slopes is \frac{1}{t_{1}}-\frac{1}{t_{2}}=-1

Therefore, the tangents are at right angles. Also, the point of intersection of  these tangents is x=at_{1}t_{2}, y=a(t_{1}+t_{2}), that is, x=-a, y=a(t_{1}+t_{2}), which clearly lies on the directrix x=-a.

(B) Proposition 2: The tangent at any point of a parabola bisects the angle between the focal distance of the point and the perpendicular on the directrix from that point. Homework !

(C) Proposition 3: The portion of a tangent to a parabola cut off between the directrix and the curve, subtends a right angle at the focus. Homework!

Problem 2:

Theorem:

Show that in general, three normals can be drawn to a given parabola from a given point, one of which is always real. Also, show that the sum of the ordinates of the feet of these co-normal points is zero.

Proof:

Let y^{2}=4ax be the given parabola. The equation of a normal to this parabola at (am^{2},-2am) is y=mx-2am-am^{2}. If it passes through a given point (\alpha, \beta) then

\beta = m\alpha -2am-am^{3}, or am^{3}+(2a-\alpha)m + \beta = 0…call this Equation (I)

which, being cubic in m, gives three values of m, say m_{1}. m_{2}, and m_{3}, and hence, three points on the parabola, the normals at which pass through (\alpha, \beta). Since the complex roots of the equation with real coefficients occur in pairs and the degree of the above equation is odd, at least one of the roots is real so there is at least one real normal to the parabola passing through the given point (\alpha, \beta).

From (I), we have

m_{1}+m_{2}+m_{3}=0 \Longrightarrow -2am_{1}-2am_{2}-2am_{3}=0 \Longrightarrow y_{1}+y_{2}+y_{3}=0, where y_{i}=-2am_{i} for i=1, 2, 3. Hence, the result.

Problem 3:

If the tangents and normals at the extremities of a focal chord of the parabola y^{2}=4ax intersect at (x_{1},y_{1}) and (x_{2},y_{2}) respectively, then show that y_{1}=y_{2}.

Answer 3:

Let P(at_{1}^{2}, 2at_{1}) and Q(at_{2}^{2},2at_{2}) be the extremities of a focal chord of the parabola y^{2}=4ax, then

t_{1}t_{2}=-1….call this Relation (I).

Next, equations of the tangents at P and Q are respectively, t_{1}y=x+at_{1}^{2} and t_{2}y=x+at_{2}^{2}

Solving these equations, we get

(t_{1}-t_{2})y=a(t_{1}^{2}-t_{2}^{2}) \Longrightarrow y=a(t_{1}+t_{2}) and

x=t_{1}a(t_{1}+t_{2})-at_{1}^{2}=at_{1}t_{2}=-a

So that x_{1}=-a and y_{1}=a(t_{1}+t_{2})….call this Relation (II).

Now, equations of the normals at P and Q are respectively

y = -t_{1}x+2at_{1}+at_{1}^{3} and y=-t_{2}x+2at_{2}+at_{2}^{3}

Solving these equations we get

-(t_{1}-t_{2})x+2a(t_{1}-t_{2})+a(t_{1}^{3}-t_{2}^{3})=0

\Longrightarrow x =2a +a(t_{1}^{2}+t_{2}^{2}+t_{1}t_{2}) = a(t_{1}^{2}+t_{2}^{2}+1) (using relation I)

and y = -t_{1}a(t_{1}^{2}+t_{2}^{2}+1) + 2at_{1}+at_{1}^{3} = -at_{1}t_{2}^{2}+at_{1}=a(t_{1}+t_{2}) (again by using relationship I),

So that x_{2}=a(t_{1}^{2}+t_{2}^{2}+1) and y_{2}=a(t_{1}+t_{2}) so that we get y_{1}=y_{2}.

Note:

Results or relations I and II should be memorized as they are frequently used in the theory and applications.

More later,

Nalin Pithwa.

 

Practice Quiz on Conic Sections (Parabola, Ellipse, Hyperbola): IITJEE Mains — basics 1

Multiple Choice Questions:

Problem 1:

A line bisecting the ordinate PN of a point P (at^{2}, 2at), t>0, on the parabola y^{2}=4ax is drawn parallel to the axis to meet the curve at Q. If NQ meets the tangent at the vertex at the point T, then the coordinates of T are

(a) (0, \frac{4}{3}at) (b) Slatex (0,2at)$ (c) (\frac{1}{4}at^{2},at) (d) (0,at)

Problem 2:

If P, Q, R are three points on a parabola y^{2}=4ax whose ordinates are in geometrical progression, then the tangents at P and R meet on

(a) the line through Q parallel to x-axis

(b) the line through Q parallel to y-axis

(c) the line through Q to the vertex

(d) the line through Q to the focus.

Problem 3:

The locus of the midpoint of the line segment joining the focus to a moving point on the parabola y^{2}=4ax is another parabola with directrix:

(a) x=-a (b) x=-a/2 (c) x=0 (d) x=a/2

Problem 4:

Equation of the locus of the pole with respect to the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1, of any tangent line to the auxiliary circle is the curve \frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}}=\lambda^{2} where

(a) \lambda^{2}=a^{2} (b) \lambda^{2}=\frac{1}{a^{2}} (c) \lambda^{2}=b^{2} (d) \lambda^{2}=\frac{1}{b^{2}}

Problem 5:

The locus of the points of the intersection of the tangents at the extremities of the chords of the ellipse x^{2}+2y^{2}=6, which touch the ellipse x^{2}+4y^{2}=4 is

(a) x^{2}+y^{2}=4 (b) x^{2}+y^{2}=6 (c) x^{2}+y^{2}=9 (d) none of these.

Problem 6:

If an ellipse slides between two perpendicular straight lines, then the locus of its centre is

(a) a parabola (b) an ellipse (c) a hyperbola (d) a circle

Problem 7:

Let P(a\sec{\theta}, b\tan{\theta}) and Q(a\sec{\phi}, b\tan{\phi}) where \theta + \phi=\pi/2, be two points on the hyperbola \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}=1. If (h,k) is the point of intersection of normals at P and Q, then k is equal to

(a) \frac{a^{2}+b^{2}}{a} (b) -(\frac{a^{2}+b^{2}}{a}) (c) \frac{a^{2}+b^{2}}{b} (d) -(\frac{a^{2}+b^{2}}{b})

Problem 8:

A straight line touches the rectangular hyperbola 9x^{2}-9y^{2}=8, and the parabola y^{2}=32x. An equation of the line is

(a) 9x+3y-8=0 (b) 9x-3y+8=0 (c) 9x+3y+8=0 (d) 9x-3y-8=0

There could be multiple answers to this question.

Problem 9:

Two parabolas C and D intersect at the two different points, where C is y=x^{2}-3 and D is y=kx^{2}. The intersection at which the  x-value is positive is designated point A, and x=a at this intersection. The tangent line l at A to the curve D intersects curve C at point B, other than A. If x-value of point B is 1(one), then what is a equal to?

Problem 10:

The triangle formed by the tangent to the parabola y=x^{2} at the point whose abscissa is x_{0}(x_{0} \in [1,2]), the y-axis and the straight line y=a_{0}^{2} has the greatest area if x_{0}= ?. Fill in the question mark!

More later,

Nalin Pithwa.

PS: I think let’s continue this focus on co-ordinate geometry of IITJEE Mains Maths for some more time.

 

Some more examples of pair of straight lines questions for IITJEE math

Problem 1:

If one of the straight lines given by the equation ax^{2}+2hxy+by^{2}=0 coincides with one of those given by a^{'}x^{2}+2h^{'}xy+b^{'}y^{2}=0, and the other lines represented by them be perpendicular, prove that \frac{ha^{'}b^{'}}{b^{'}-a^{'}}=\frac{h^{'}ab}{b-a}=\frac{1}{2}\sqrt{aa^{'}bb^{'}}

Solution 1:

Let the lines represented by ax^{2}+2hxy+by^{2}=0 be y=m_{1}x and y=m_{2}x, so that

m_{1}+m_{2}=\frac{2h}{b} and m_{1}m_{2}=\frac{a}{b}…call this equation (A)

The lines represented by a^{'}x^{2}+2h^{'}xy+b^{'}y^{2}=0 are y=m_{1}x and y=-\frac{1}{m_{2}}x_{1} so that

m_{1}-\frac{1}{m_{2}}=-\frac{2h^{'}}{b^{'}} and m_{1}(-\frac{1}{m_{2}})=\frac{a^{'}}{b^{'}}…call this equation (B)

From (A) and (B), we get m_{1}m_{2}(-\frac{m_{1}}{m_{2}})=\frac{aa^{'}}{bb^{'}}

which in turn \Longrightarrow m_{1}^{2}=\frac{aa^{'}}{bb^{'}},

which in turn \Longrightarrow m_{1}=\sqrt{(-\frac{aa^{'}}{bb^{'}})}

and from (A), again, we get \sqrt{-\frac{aa^{'}}{bb^{'}}}-\sqrt{-\frac{aa^{'}}{ba^{'}}}=-\frac{2h}{b}

and this in turn \Longrightarrow \sqrt{(-aa^{'}bb^{'})}(\frac{1}{bb^{'}}-\frac{1}{a^{'}b})=-\frac{2h}{b}, that is, it \Longrightarrow \frac{1}{2}\sqrt{(-aa^{'}bb^{'})}=\frac{ha^{'}b^{'}}{b^{'}-a^{'}}.

Similarly, from (B), we get \frac{1}{2}\sqrt{(-aa^{'}bb^{'})}=\frac{h^{'}ab}{b-a}. Hence, the required result follows.

Problem 2:

If the equation ax^{2}+2hxy+by^{2}+2gx+2fy+c=0 represents two straight lines, prove that the equation of the third pair of straight lines passing through the points where these meet the axes xy=0 is c(ax^{2}+2hxy+by^{2}+2gx+2fy+c)+4(fg+ch)xy=0.

Solution 2:

Equation of the lines passing through the intersection of  the given lines and axes is

ax^{2}+2hxy+by^{2}+2gx+2fy+c+2xy \lambda =0, or

ax^{2}+2(h+\lambda)xy+by^{2}+2gx+2fy+c=0

Since it represents a pair of straight lines

abc+2fg(h+\lambda)-af^{2}-bg^{2}-c(h+\lambda)^{2}=0,

\Longrightarrow abc+2fgh-af^{2}-bg^{2}-ch^{2}+2fg\lambda -2ch\lambda -c\lambda^{2}=0

\Longrightarrow \lambda = \frac{2(fg-ch)}{c} (since abc+2fgh-af^{2}-bg^{2}-ch^{2}=0).

Hence, the required equation of the lines is c(ax^{2}+2hxy+by^{2}+2gx+2fy+c)+4(fg+ch)xy=0

Problem 3:

Show that the equation \sqrt{3}x^{2}-(4+\sqrt{3})x^{2}y+(4+\sqrt{3})xy^{2}-\sqrt{3}y^{2}=0 represents three straight lines through the origin such that one of them bisects the angle between the other two. Also find the equation of  the lines perpendicular to the given lines through the origin.

Solution 3:

The given equation can be written as \sqrt{3}(x^{3}-y^{3})-(4+\sqrt{3})xy(x-y)=0, or

(x-y)(\sqrt{3}(x^{2}+xy+y^{2}))-(4+\sqrt{3})xy=0, or

(x-y)(\sqrt{3}x^{2}-4xy+\sqrt{3}y^{2})=0, or

(x-y)(\sqrt{3}x-y)(x-\sqrt{3}y)=0,or

x-y=0, \sqrt{3}x-y=0, and x-\sqrt{3}y=0

which gives three straight lines passing through the origin with slopes 45 degrees, 60 degrees, and 30 degrees respectively showing that x-y=0 bisects the angles between the other two.

Now, equations of the lines through the origin perpendicular to these lines are

x+y=0, x+\sqrt{3}y=0, \sqrt{3}x+y=0 so their joint equation is given by

(x+y)(x+\sqrt{3}y)(\sqrt{3}x+y)=0

or, \sqrt{3}x^{3}+(4+\sqrt{3})x^{2}y+(4+\sqrt{3})xy^{2}+\sqrt{3}y^{3}=0

That’s all, folks !

Nalin Pithwa.