Question

If Latus rectum of parabola $y^2 = 4kx$ and ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ coincide then the value of $e^2 + 2\sqrt{2}$ is, where $e$ is eccentricity of ellipse

• A. 2
• B. 3
• C. 4
• D. 3/2

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
"Latus rectum coincide" implies they have the same focus and the same line segment length.
For the parabola $y^2 = 4kx$, focus is $(k, 0)$ and length of LR is $4k$.
For the ellipse, focus is $(ae, 0)$ and length of LR is $\frac{2b^2}{a}$.
Step 2: Key Formula or Approach:
1. Set focus $k = ae$.
2. Set LR length $4k = \frac{2b^2}{a}$.
3. Use the ellipse relation $b^2 = a^2(1 - e^2)$.
Step 3: Detailed Explanation:
1. Equating lengths:
$4k = \frac{2b^2}{a} \implies 4(ae) = \frac{2b^2}{a} \implies 2a^2e = b^2$.
2. Substitute $b^2 = a^2(1 - e^2)$:
$2a^2e = a^2(1 - e^2) \implies 2e = 1 - e^2 \implies e^2 + 2e - 1 = 0$.
3. Solve the quadratic for $e$ (using $e>0$):
$e = \frac{-2 \pm \sqrt{4 + 4}}{2} = -1 \pm \sqrt{2}$.
Since $0<e<1$, we take $e = \sqrt{2} - 1$.
4. Calculate $e^2$:
$e^2 = (\sqrt{2} - 1)^2 = 2 + 1 - 2\sqrt{2} = 3 - 2\sqrt{2}$.
5. Find the requested value:
$e^2 + 2\sqrt{2} = (3 - 2\sqrt{2}) + 2\sqrt{2} = 3$.
Step 4: Final Answer:
The value is 3.