Question

A chord is drawn passing through $P\left(2 , \, 2\right)$ on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$ such that it intersects the ellipse at points $A$ and $B$ . Then the maximum value of $PA\cdot PB$ is equal to $\frac{a}{b}$ where $a,b$ are the least common multiple then the value of $a+b$ is

Answer 1

The chord is passing through the point $P(2,2)$.
So, the equation of chord in parametric form will be: $\frac{x-2}{\cos \theta}=\frac{y-2}{\sin \theta}=r$.
Now, on solving the equation of the chord with
equation of ellipse will give $r$ of the points $A$ and $B$
$\therefore \frac{(r \cos \theta+2)^{2}}{25}+\frac{(r \sin \theta+2)^{2}}{16}=1$
$\Rightarrow 16(r \cos \theta+2)^{2}+25(r \sin \theta+2)^{2}=400$
$\Rightarrow 16 r^{2} \cos ^{2} \theta+64 r \cos \theta+64+25 r^{2} \sin ^{2} \theta+100+100 r \sin \theta=400$
$\Rightarrow r^{2}\left(16(\cos )^{2} \theta+25(\sin )^{2} \theta\right)+r(64 \cos \theta+100 \sin \theta)-236=0$,
which is quadratic equation in $r$.
$\Rightarrow\left|r_{1} r_{2}\right|=P A \cdot P B=\left|\frac{-236}{16 \cos ^{2} \theta+25 \sin ^{2} \theta}\right|$
$=\left|\frac{236}{16+9 \sin ^{2} \theta}\right|$
Since, range of $\sin \theta \in[-1,1]$.
Maximum value occur when denominator is minimum.
Therefore, the maximum value of $P A \cdot P B=\frac{236}{16}=\frac{59}{4}$.