Ellipse Question - None

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 $(P A) \cdot(P B)$ is maximum. Then $5\left(P A^{2}+P B^{2}\right)$ is equal to:

Accepted Answer

Any point $Q$ on the line passing through $P(\sqrt{5}, \sqrt{5})$ can be expressed in parametric form as:
$Q(\sqrt{5}+r \cos \theta, \sqrt{5}+r \sin \theta)$
where $r$ is the distance from $P$ to $Q$, and $\theta$ is the angle the line makes with the $x$-axis. Substitute $Q$ into the ellipse equation $\frac{x^{2}}{36}+\frac{y^{2}}{25}=1$:
$25(\sqrt{5}+r \cos \theta)^{2}$$+36(\sqrt{5}+r \sin \theta)^{2}=900$
Expanding and simplifying, we obtain:
$r^{2}\left(25 \cos ^{2} \theta+36 \sin ^{2} \theta\right)+$ $2 \sqrt{5} r(25 \cos \theta+36 \sin \theta)-595=0$
The distances $P A$ and $P B$ correspond to the absolute values of the roots $r_{1}$ and $r_{2}$ of the quadratic equation in $r$. The product of the roots is:
$P A \cdot P B =\left|\frac{-595}{25 \cos ^{2} \theta+36 \sin ^{2} \theta}\right|$
$=\frac{595}{25+11 \sin ^{2} \theta}$
To maximize this product, the denominator $25+11 \sin ^{2} \theta$ must be minimized. This occurs when $\sin ^{2} \theta=0$, i.e., when the line is horizontal $(\theta=0)$. For a horizontal line $(y=\sqrt{5})$, substitute into the ellipse equation:
$\frac{x^{2}}{36}+\frac{5}{25}=1 \Rightarrow \frac{x^{2}}{36}=\frac{4}{5} \Rightarrow x= \pm \frac{12}{\sqrt{5}}$
Thus, the points of intersection are:
$A\left(-\frac{12}{\sqrt{5}}, \sqrt{5}\right)$ and $B\left(\frac{12}{\sqrt{5}}, \sqrt{5}\right)$
$P A^{2}+P B^{2}=\left(\sqrt{5}+\frac{12}{\sqrt{5}}\right)^{2}+\left(\sqrt{5}-\frac{12}{\sqrt{5}}\right)^{2}$
$=2\left(5+\frac{144}{5}\right)=\frac{338}{5}$
$\therefore 5\left(P A^{2}+P B^{2}\right)=5 \cdot \frac{338}{5}=338$

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