Question

Three distinct chords drawn from $(\alpha ,0)$ to the ellipse $x^2+2y^2=1.$ If these chords are bisected by the parabola $y^2=4x,$ then $\left[\alpha \right]=$
(where $[.]$ denotes greatest integer function)

Answer 1

Let the middle point of chord be $(t^2,2t)$

Midpoint of chord must lie inside the ellipse $x^2+2y^2=1,$ thus
$\Rightarrow t^4+8t^2-1<0$
$\Rightarrow t^2\in [0,\sqrt{17}-4)\dots \left(1\right)$

Also equation of the chord, whose mid point is $(t^2,2t)$ is given by $T=S_1$
$\Rightarrow t^{2}x+4ty=t^4+8t^2$

If it passes through the point $(\alpha ,0),$ then
$\Rightarrow \alpha t^2=t^4+8t^2$
$\Rightarrow t^4+(8-\alpha )t^2=0$
$\Rightarrow t^2=0$ or $t^2=\alpha -8$
So for three distinct chord three distinct values of $t$ should exists
$\therefore t=0,\pm \sqrt{\alpha -8}$
Hence $\alpha =t^2+8\dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$ we get
$\Rightarrow \alpha \in [8,4+\sqrt{17})\therefore \left[\alpha \right]=8$