Question

A tangent to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ with centre $C$ meets its director circle at $P$ and $Q.$ Then the product of the slopes of $CP$ and $CQ,$ is :

• A. $\frac{9}{4}$
• B. $\frac{-4}{9}$
• C. $\frac{2}{9}$
• D. $-\frac{1}{4}$

Answer 1

The correct option is B $\frac{-4}{9}$

Director circle:

$x^2+y^2=9+4=13$

$\Rightarrow x^2+y^2=13$

Equation of tangent to ellipse with point of contact $(3\cos \theta ,2\sin \theta )$

$\Rightarrow T=0$

$\Rightarrow \frac{x\cos \theta }{3}+\frac{y\sin \theta }{2}=1$

Homogenissing with director circle gives :

$CP\times CQ$ equation

$\Rightarrow x^2+y^2=13(\frac{x\cos \theta }{3}+\frac{y\sin \theta }{2}{)}^2$

$\Rightarrow x^2+y^2=13(\frac{x^2{\cos }^2\theta }{9}+\frac{y^2{\sin }^2\theta }{4}+\frac{xy}{3}\cos \theta .\sin \theta )$

$\Rightarrow x^2(\frac{13}{9}{\cos }^2\theta -1)+y^2(\frac{13}{4}{\sin }^2\theta -1)+\frac{13xy}{3}\cos \theta .\sin \theta =0$

Product of slopes of $CP$ and $CQ$

$=\frac{{\text{Coff of x}}^2}{{\text{Coff of y}}^2}$

$=\frac{\frac{13}{9}{\cos }^2\theta -1}{\frac{13}{4}{\sin }^2\theta -1}$

$=\left(\frac{13{\cos }^2\theta -9}{13{\sin }^2\theta -4}\right)\frac{4}{9}$

$=\left(\frac{13(1-{\sin }^2\theta )-9}{13{\sin }^2\theta -4}\right)\frac{4}{9}$

$=\left(\frac{4-13{\sin }^2\theta }{13{\sin }^2\theta -4}\right)\frac{4}{9}$

$=\frac{-4}{9}\left(\frac{13{\sin }^2\theta -4}{13{\sin }^2\theta -4}\right)$

$=\frac{-4}{9}$

$\therefore B)$ Answer.