## 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.

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