Question

If a straight line through the point $P(\lambda ,2)$ where $\lambda \ne 0$, meets the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ at $A$ and $D$ and meets the coordinate axis at $B$ and $C$ such that $PA\cdot PD=PB\cdot PC,$ then range of $\lambda$ is
(correct answer + 2, wrong answer - 0.50)

• A. $(-\infty ,-6]\cup [6,\infty )$
• B. $(-\infty ,-12]\cup [12,\infty )$
• C. $[6,\infty )$
• D. $[12,\infty )$

Answer 1

The correct option is A $(-\infty ,-6]\cup [6,\infty )$

Given : $\frac{x^2}{9}+\frac{y^2}{4}=1$

Any point on the line passing through $P(\lambda ,2)$ is $(\lambda +r\cos \theta ,2+r\sin \theta )$
Intersection with ellipse, we get
$\frac{(\lambda +r\cos \theta {)}^2}{9}+\frac{(2+r\sin \theta {)}^2}{4}=1\Rightarrow (4{\cos }^2\theta +9{\sin }^2\theta )r^2+(8\lambda \cos \theta +36\sin \theta )r+4{\lambda }^2=0\Rightarrow (5{\sin }^2\theta +4)r^2+(8\lambda \cos \theta +36\sin \theta )r+4{\lambda }^2=0$
The roots are $r_1,r_2$, so
$r_1{r}_2=\frac{4{\lambda }^2}{5{\sin }^2\theta +4}\Rightarrow PA\cdot PD=\frac{4{\lambda }^2}{5{\sin }^2\theta +4}$

Intersection with $x-$ axis
$2+r_3\sin \theta =0\Rightarrow r_3=-\frac{2}{\sin \theta }$
Intersection with $y-$ axis
$\lambda +r_4\cos \theta =0\Rightarrow r_4=-\frac{\lambda }{\cos \theta }\Rightarrow PB\cdot PC=r_3{r}_4=\frac{2\lambda }{\sin \theta \cos \theta }$


$PA\cdot PD=PB\cdot PC\Rightarrow \frac{4{\lambda }^2}{5{\sin }^2\theta +4}=\frac{2\lambda }{\sin \theta \cos \theta }\Rightarrow \lambda \sin 2\theta =\frac{5(1-\cos 2\theta )}{2}+4\Rightarrow 2\lambda \sin 2\theta +5\cos 2\theta =13\Rightarrow -1\le \frac{13}{\sqrt{4{\lambda }^2+25}}\le 1\Rightarrow 0\le \frac{169}{4{\lambda }^2+25}\le 1\Rightarrow 144\le 4{\lambda }^2\Rightarrow {\lambda }^2\ge 36\therefore \lambda \in (-\infty ,-6]\cup [6,\infty )$