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

Answer: 63

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 $B$
$\therefore \frac{(r\cos \theta +2{)}^2}{25}+\frac{(r\sin \theta +2{)}^2}{16}=1$
$\Rightarrow 16(r\cos \theta +2{)}^2+25(r\sin \theta +2{)}^2=400$
$\Rightarrow 16r^2{\cos }^2\theta +64r\cos \theta +64+25r^2{\sin }^2\theta +100+100r\sin \theta =400$
$\Rightarrow r^2\left(16(\cos {)}^2\theta +25(\sin {)}^2\theta \right)+r(64\cos \theta +100\sin \theta )-236=0$,
which is quadratic equation in $r$.
$\Rightarrow \left|r_1{r}_2\right|=PA\cdot PB=\left|\frac{-236}{16{\cos }^2\theta +25{\sin }^2\theta }\right|$
$=\left|\frac{236}{16+9{\sin }^2\theta }\right|$
Since, range of $\sin \theta \in [-1,1]$.
Maximum value occur when denominator is minimum.
Therefore, the maximum value of $PA\cdot PB=\frac{236}{16}=\frac{59}{4}$.