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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?

  • Do you mean "two real roots for $\alpha > 8$"? – little o Apr 18 '19 at 08:18
  • @Dbchatto for $\alpha<8$ roots for $t^2$ will be imaginary –  Apr 18 '19 at 08:43
  • Yeah! I know that. You wrote $\alpha > 0$ in the previous edit. That's why I was confirming. – little o Apr 18 '19 at 08:45
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    The equation of the chord passing through $(t^2,2t)$ and $(\alpha,0)$ is $$2 tx + (\alpha-t^2)y = 2 \alpha t.$$ Try to show that it would not intersect the given ellipse if $\alpha \geq 4 + \sqrt {17}.$ – little o Apr 18 '19 at 09:01

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