A line passing through the point P(√5,√5) intersects the ellipse x^236+y^225=1 at A and B …

Question

A line passing through the point $P(\sqrt{5},\sqrt{5})$ intersects the ellipse $\frac{{x}^{2}}{36}+\frac{{y}^{2}}{25}=1$ at $A$ and $B$ such that $(PA)\cdot (PB)$ is maximum. Then $5(P{A}^{2}+P{B}^{2})$ is equal to :

Accepted Answer

$$\begin{aligned}
& \frac{x-\sqrt{5}}{\cos \theta}=\frac{y-\sqrt{5}}{\sin \theta}=r \
& x=r \cos \theta+\sqrt{5} \
& y=r \sin \theta+\sqrt{5} \
& 25 x^2+36 y^2=900,(x, y) \text { lie on } E \
& r^2\left[25 \cos ^2 \theta+36 \sin ^2 \theta\right]+r[50 \sqrt{5} \cos \theta+72 \sqrt{5} \sin \theta] \
& \quad+25[5]+36[5]=900 \
& r_1 r_2=\frac{900-305}{25+11 \sin ^2 \theta},\left(r_1 r_2\right)_{\max }=\frac{595}{25}=\frac{119}{5} \
& \left|r_1 \cdot r_2\right|_{\max }=\left(\frac{595}{25}\right)=\left(\frac{119}{5}\right) \text { at } \theta=0
\end{aligned}$$

$$\begin{aligned}
& \frac{x^2}{36}+\frac{y^2}{25}=1 \
& \text { At } y=\sqrt{5} \
& \frac{x^2}{36}+\frac{1}{5}=1 \
& \Rightarrow \frac{x^2}{36}=\frac{4}{5} \Rightarrow x= \pm \frac{12}{\sqrt{5}} \
& P A^2+P B^2=(P A+P B)^2-2 P A \cdot P B=\left(\frac{24}{\sqrt{5}}\right)^2-2 \cdot \frac{119}{5} \
& 5\left(P A^2+P B^2\right)=5\left(\frac{24^2}{5}-\frac{2.119}{5}\right)=24^2-238 \
& =338 .
\end{aligned}$$

Solutions · 1