Question

Consider the parabola \(P : y^2 = 4kx\) and the ellipse \(E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Let the line segment joining the points of intersection of \(P\) and \(E\), be their latus rectum. If the eccentricity of \(E\) is \(e\), then \(e^2 + 2\sqrt{2}\) is equal to ____.

Answer 1

Correct Answer: 3

Step 1: Understanding the Concept:
The latus rectum of the parabola $y^2 = 4kx$ is the line $x = k$. The length of this latus rectum is $4k$. For this to be the common latus rectum, the points of intersection must be $(k, 2k)$ and $(k, -2k)$.

Step 2: Key Formula or Approach:
1. Points $(k, \pm 2k)$ lie on the ellipse: $\frac{k^2}{a^2} + \frac{4k^2}{b^2} = 1$.
2. For an ellipse, the latus rectum length is $\frac{2b^2}{a}$. Here, $4k = \frac{2b^2}{a} \implies b^2 = 2ak$.
3. Eccentricity $e^2 = 1 - \frac{b^2}{a^2}$.

Step 3: Detailed Explanation:
1. Substitute $b^2 = 2ak$ into the ellipse equation:
$\frac{k^2}{a^2} + \frac{4k^2}{2ak} = 1 \implies \frac{k^2}{a^2} + \frac{2k}{a} - 1 = 0$.
2. Let $t = k/a$. Then $t^2 + 2t - 1 = 0$.
3. Solving for $t$ (positive): $t = \frac{-2 + \sqrt{4 + 4}}{2} = \sqrt{2} - 1$.
4. Now, $e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{2ak}{a^2} = 1 - 2(\frac{k}{a}) = 1 - 2t$.
5. $e^2 = 1 - 2(\sqrt{2} - 1) = 1 - 2\sqrt{2} + 2 = 3 - 2\sqrt{2}$.
6. Calculate $e^2 + 2\sqrt{2} = (3 - 2\sqrt{2}) + 2\sqrt{2} = 3$.

Step 4: Final Answer:
The value is 3.