A line passing through the point P(sqrt{5}, s | vedclass

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(D) The given ellipse is $\frac{x^2}{36} + \frac{y^2}{25} = 1$.Any point on the line $AB$ passing through $P(\sqrt{5}, \sqrt{5})$ can be represented a

Vedclass · Accepted Answer

$338$

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(D) The given ellipse is $\frac{x^2}{36} + \frac{y^2}{25} = 1$.Any point on the line $AB$ passing through $P(\sqrt{5}, \sqrt{5})$ can be represented as $Q(\sqrt{5} + r \cos 	heta, \sqrt{5} + r \sin 	heta)$.Substituting this into the equation of the ellipse,we get:$25(\sqrt{5} + r \cos 	heta)^2 + 36(\sqrt{5} + r \sin 	heta)^2 = 900$Expanding and simplifying,we obtain:$r^2(25 \cos^2 	heta + 36 \sin^2 	heta) + 2\sqrt{5}r(25 \cos 	heta + 36 \sin 	heta) - 595 = 0$Here,$|r| = PA, PB$. The product of the roots is $PA \cdot PB = \left| \frac{-595}{25 \cos^2 	heta + 36 \sin^2 	heta} ight| = \frac{595}{25 + 11 \sin^2 	heta}$.This product is maximum when the denominator is minimum,i.e.,when $\sin^2 	heta = 0$.This implies the line $AB$ is parallel to the $x$-axis,so $y = \sqrt{5}$.Substituting $y = \sqrt{5}$ into the ellipse equation: $\frac{x^2}{36} + \frac{5}{25} = 1$ $\Rightarrow \frac{x^2}{36} = \frac{4}{5}$ $\Rightarrow x^2 = \frac{144}{5}$ $\Rightarrow x = \pm \frac{12}{\sqrt{5}}$.The points are $A(-\frac{12}{\sqrt{5}}, \sqrt{5})$ and $B(\frac{12}{\sqrt{5}}, \sqrt{5})$.$PA^2 = (\sqrt{5} - (-\frac{12}{\sqrt{5}}))^2 + (\sqrt{5} - \sqrt{5})^2 = (\frac{5+12}{\sqrt{5}})^2 = \frac{289}{5}$.$PB^2 = (\sqrt{5} - \frac{12}{\sqrt{5}})^2 + (\sqrt{5} - \sqrt{5})^2 = (\frac{5-12}{\sqrt{5}})^2 = \frac{49}{5}$.$PA^2 + PB^2 = \frac{289+49}{5} = \frac{338}{5}$.Therefore,$5(PA^2 + PB^2) = 338$.

$A$ 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 $5(PA^2 + PB^2)$ is equal to:

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