Find the values of $\alpha$ for which three distinct chords from $(\alpha, 0)$ to the ellipse $x^2 + 2y^2=1$ are bisected by the parabola $y^2=4x$
Parametric form of the parabola is $(t^2,2t)$ also this is the of the chord so for $T=S_1$ (equation of a chord bisected at $(x_1 , y_1 )$) $(x_1=t^2, y_1=2t)$
Substituting $(x_1=t^2, y_1=2t)$ for the ellipse we get the equation of the chord as. $$t^2x + 4ty=t^4 +8t^2$$
$(\alpha , 0)$ lies on this chord.
So $$t^2(\alpha-8-t^2)=0$$ will have one root at $t=0$ and two real roots for $\alpha>8$ however the range for alpha is given as $(8,4+ \sqrt 17)$
Why did I not get the upper limit in my solution?