Question

Line through $P(a,2)$ meets the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ at A and D and meets the coordinate axes at B and C so that PA, PB, PC, PD are in G.P., then possible values of a can be?

• A. $5$
• B. $8$
• C. $10$
• D. $-7$

Answer 1

Answer: (B), (C)

$8$
$10$

The equation of line is

$\begin{array}{c}y-2=\left(x-a\right)\tan \theta \\ TheP\left(A\right)=\left[a+r\cos \theta ,2+r\sin \theta \right]\end{array}$

Where, r is distance between $P\left(A\right)$ & $P(a,2)$

Solving equation of straight line by putting $x=0$

and $y=0$, we get

$\begin{array}{c}P\left(B\right)=\left(a-2\cot \theta ,0\right)\\ P\left(C\right)=\left(0,2-a\tan \theta \right)\\ Distance,PB=\sqrt{{\left(2\left(\cos t\theta \right)\right)}^2+4}=\sqrt{4{\cot }^2+4}\\ =2\cos ec\theta \\ PC=\sqrt{a^2+{\left(a\tan \theta \right)}^2}=a\sec \theta \end{array}$

Similarly, we get, PD and PA by solving straight line and ellipse equation.

$\begin{array}{c}PA.PD=PB.PC\\ \frac{PB}{PA}=\frac{PD}{PC}\\ \frac{4a^2}{4+5{\sin }^2\theta }=\frac{2a}{\sin \theta \cot \theta }\\ 4a=\frac{2\left(4+5{\sin }^2\theta \right)}{\sin \theta \cos \theta }\\ a=\frac{13-5\cos 2\theta }{2\sin 2\theta }\\ =\frac{\left[13-5\left(1-{\tan }^2\theta \right)\left(1+{\tan }^2\theta \right)\right]}{\left[4\tan \theta /\left(1+{\tan }^2\theta \right)\right]}\end{array}$

On Solving, we get quadratic equation of $\tan \theta$

Solving, putting $D=0,$ we get

$\begin{array}{c}{a}^2>36\\ \Rightarrow a>6\\ \therefore BandC\end{array}$