Question

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

Answer 1

Answer: 0002

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