Monthly Archives: September 2017

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.