Question

Find the number of integral values of parameter a for which three chords of the ellipse $\frac{ x ^2}{2 a ^2}+\frac{ y ^2}{ a ^2}=1$ (other than its diameter) passing through the point $P \left(11 a , \frac{- a ^2}{4}\right)$ are bisected by the parabola $y^2=4 a x$

Answer 1

Any point on the parabola $y^2=4 ax$ is $\left( at ^2, 2 at \right)$. Equation of chord of the ellipse $\frac{x^2}{2 a^2}+\frac{y^2}{a^2}=1$, whose mid-point is ( $\left.a t^2, 2 a t\right)$ is $\frac{x \cdot a t^2}{2 a^2}+\frac{y \cdot 2 a t}{a^2}=\frac{a^2 t^4}{2 a^2}+\frac{4 a^2 t^2}{a^2}$
$\Rightarrow tx +4 y = at ^3+8 at (\because t \neq 0)$
As it passes through $\left(11 a , \frac{- a ^2}{4}\right)$,
$\Rightarrow 11 a t-4\left(\frac{ a ^2}{4}\right)=a t ^3+8 at $
$\Rightarrow a t ^3-3 at + a ^2=0 \Rightarrow t ^3-3 t + a =0 ( a \neq 0)$
Now, three chords of the ellipse will be bisected by the parabola if the equation (1) has three real and distinct roots.
$\text { Let } f(t)=t^3-3 t+a $
$ f^{\prime}(t)=3 t^2-3=0 \Rightarrow t= \pm 1$
So, $ f(1) f(-1)<0$
$\Rightarrow a \in(-2,2)$
But $ a \neq 0$, so $a \in(-2,0) \cup(0,2)$
Hence number of integral values of a are 2 (i.e. -1 and 1.)