Question

If $Ax+By=5$ is a normal at a point $P$ on the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ whose eccentric angle is $\frac{\pi }{4}$, then the value of $(A+B{)}^2$ is

Answer 1

Given: $\frac{x^2}{9}+\frac{y^2}{4}=1$
Here,
$a=3,b=2$
Using parametric form of normal:
$\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2\Rightarrow \frac{3x}{\cos \theta }-\frac{2y}{\sin \theta }=5$

For $\theta =\frac{\pi }{4}$ the equation of normal is
$\Rightarrow 3\sqrt{2}x-2\sqrt{2}y=5$
Comparing with $Ax+By=5$, we get
$A=3\sqrt{2},B=-2\sqrt{2}\therefore (A+B{)}^2=2$