Consider the circle C : x^2 + y^2 = 4 and the parabola P : y^2 = 8x. If the set of all values of α,

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 \(\mathrm{C}\) on three distinct lines passing through the point \((\alpha, 0)\) are bisected by the parabola P is the interval \((p, q)\), then \((2 q-p)^{2}\) is equal to _________.

Answer 1

Correct answer is : 80 

\( \begin{aligned} & T=S_1 \end{aligned} \) 

\( \begin{aligned} x_1+y_1 y_1=x_1^2+y_1^2 \end{aligned} \) 

It passes through \((\alpha, 0)\)n 

\( \begin{aligned} & \therefore \alpha \mathrm{x}_1=\mathrm{x}_1^2+\mathrm{y}_1^2 \end{aligned} \)

\( \begin{aligned} \alpha\left(2 \mathrm{t}^2\right)=4 \mathrm{t}^4+16 \mathrm{t}^2 \quad\left(\mathrm{x}_1=2 \mathrm{t}^2, \mathrm{y}_1=4 \mathrm{t}\right) \\ \end{aligned} \)

\( \begin{aligned} & \alpha=2 \mathrm{t}^2+8 \\ \end{aligned} \)

\( \begin{aligned} & \mathrm{t}^2=\frac{\alpha-8}{2} \\ \end{aligned} \) 

\( \Rightarrow \alpha>8\) 

 Also,

\(4t^2+16t^2-4<0\) (point lies inside the circle)

\(t^2=-2+\sqrt{5}\)

\(\alpha=4+2\sqrt5\)

\(\therefore\alpha\in(8,4+2\sqrt 5)\)

\(\therefore (2q-p)^2=80\)