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

Correct Answer: 80

\[ T = S_1 \] \[ x_1 + y_1 y_1 = x_1^2 + y_1^2 \] It passes through \( (\alpha, 0) \), \[ \therefore \alpha x_1 = x_1^2 + y_1^2 \] \[ \alpha (2t^2) = 4t^4 + 16t^2 \quad \left( x_1 = 2t^2, y_1 = 4t \right) \] \[ \alpha = 2t^2 + 8 \] \[ t^2 = \frac{\alpha - 8}{2} \] \[ \Rightarrow \alpha > 8 \] Also, \[ 4t^2 + 16t^2 - 4 < 0 \quad \text{(point lies inside the circle)} \] \[ t^2 = -2 + \sqrt{5} \] \[ \alpha = 4 + 2\sqrt{5} \] \[ \therefore \alpha \in (8, 4 + 2\sqrt{5}) \] \[ \therefore (2q - p)^2 = 80 \]

Answer 2

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:\[ (2q - p)^2 = 80 \]