Question

Consider the circle $C : x^2 + y^2 = 4$ and the parabola $P : y^2 = 8x$. 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 parabola $P$, is the interval $(p, q)$, then $(2q - p)^2$ is equal to ________ .

Answer 1

Step 1. The equation of the circle is given as $T = S_1$, where:

$xx_1 + yy_1 = x_1^2 + y_1^2$

Step 2. Using the symmetry condition of the parabola and circle intersection:

$\alpha x_1 = x_1^2 + y_1^2$

Step 3. Substituting and simplifying:

$\alpha(2t^2) = 4t^4 + 16t^2$

Step 4. Rearranging:

$\alpha = 2t^2 + 8$

Step 5. Further refinement leads to:

$\frac{\alpha - 8}{2} = t^2$

Step 6. To ensure three distinct solutions, the discriminant condition for the quadratic is:

$4t^4 + 16t^2 - 4 < 0$

Solving this gives:

$t^2 = -2 \pm \sqrt{5}$

Step 7. Substituting back, $\alpha$ lies in the interval:

$\alpha \in (8, 4 + 2\sqrt{5})$

Step 8. Hence, the values of $p$ and $q$ are:

$p = 8, \quad q = 4 + 2\sqrt{5}$

Step 9. Finally, the value of $(2q - p)^2$ is:$(2q - p)^2 = 80$