Consider the circle C: x^2+y^2=4 and the parabola P: y^2=8 x.

Question

Consider the circle $C: x^2+y^2=4$ and the parabola $P: y^2=8 x$. If the set of all values of $$\alpha$$, for which three chords of the circle $C$ on three distinct lines passing through the point $$(\alpha, 0)$$ are bisected by the p...

Answer: 80

Accepted Answer

Chord with the middle point $(\alpha ,0)$

$\begin{gathered} \Rightarrow T={S}_{1} \ \Rightarrow y{y}_{1}-4(x+{x}_{1})={y}_{2}^{1}-8{x}_{1} \ \Rightarrow -4(x+\alpha )=0-8\alpha \ \Rightarrow x+\alpha =2\alpha \Rightarrow x=\alpha \end{gathered}$

For circle chord with $(2{t}^{2},4t)$ as mid point

$\begin{gathered} \Rightarrow \, T={S}_{1} \ \Rightarrow \, x{x}_{1}+y{y}_{1}-4={x}_{2}^{1}+{y}_{2}^{1}-4 \ \Rightarrow \, 2{t}^{2}x+4ty=4{t}^{4}+16{t}^{2} \end{gathered}$

Passes through $(\alpha ,0)$

$\begin{gathered} \Rightarrow 2{t}^{2}\alpha =4{t}^{4}+16{t}^{2} \ \Rightarrow 2\alpha =4{t}^{2}+16\Rightarrow \alpha =2{t}^{2}+8={x}_{0}+8 \ {x}^{2}+{y}^{2}=4{y}^{2}=8x \ \Rightarrow {x}^{2}+8x-4=0\Rightarrow {x}_{0}=\frac{-8+\sqrt{80}}{2} \ \Rightarrow p=8q=4+\frac{\sqrt{80}}{2}\Rightarrow (2q-p{)}^{2}=80 \end{gathered}$

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