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 problem states that the line segment connecting the intersection points of the parabola \(P\) and the ellipse \(E\) serves as the latus rectum for BOTH curves.
This implies that they share the exact same latus rectum, meaning their foci coincide and the lengths of their latus rectums are equal.
Step 2: Key Formula or Approach:
For the parabola \(y^2 = 4kx\):
Focus is at \((k, 0)\).
The ends of its latus rectum are \((k, 2k)\) and \((k, -2k)\).
For the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\):
Focus is at \((ae, 0)\).
The ends of its latus rectum are \((ae, \frac{b^2}{a})\) and \((ae, -\frac{b^2}{a})\).
Equating the foci and the y-coordinates of the extremities gives us the necessary equations.
Step 3: Detailed Explanation:
Since the foci coincide:
\(ae = k\) \quad (Equation 1).
Since the lengths of the semi-latus rectums are equal:
\(\frac{b^2}{a} = 2k\) \quad (Equation 2).
Substitute Equation 1 into Equation 2:
\(\frac{b^2}{a} = 2(ae) \implies b^2 = 2a^2 e\).
We also know the standard relation for an ellipse: \(b^2 = a^2(1 - e^2)\).
Equating the two expressions for \(b^2\):
\(a^2(1 - e^2) = 2a^2 e\).
Since \(a \neq 0\), we divide by \(a^2\):
\(1 - e^2 = 2e \implies e^2 + 2e - 1 = 0\).
Solving this quadratic equation for \(e\):
\(e = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}\).
Since eccentricity \(e>0\) for an ellipse, we take the positive root:
\(e = \sqrt{2} - 1\).
Step 4: Final Answer:
We are required to find the value of \(e^2 + 2\sqrt{2}\).
First, calculate \(e^2\):
\(e^2 = (\sqrt{2} - 1)^2 = 2 + 1 - 2\sqrt{2} = 3 - 2\sqrt{2}\).
Now substitute this into the given expression:
\(e^2 + 2\sqrt{2} = (3 - 2\sqrt{2}) + 2\sqrt{2} = 3\).