Question

Let $P$ be a parabola with vertex $(2, 3)$ and directrix $2x + y = 6$. Let an ellipse $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $a > b$, of eccentricity $\frac{1}{\sqrt{2}}$ pass through the focus of the parabola $P$. Then the square of the length of the latus rectum of $E$ is

• A. $\frac{385}{8}$
• B. $\frac{347}{8}$
• C. $\frac{512}{25}$
• D. $\frac{656}{25}$

Answer 1

Answer: (D)

$\frac{656}{25}$

Answer 2

Find the focus of the parabola. The equation of the directrix is:

$2x + y = 6$

The vertex of the parabola is $(2, 3)$. The equation of a parabola with vertex $(h, k)$ and directrix

$Ax + By + C = 0 \text{ has focus at: } \left( h + \frac{A}{\sqrt{A^2 + B^2}}, k + \frac{B}{\sqrt{A^2 + B^2}} \right)$

For our parabola:

$A = 2, \quad B = 1, \quad C = -6, \quad h = 2, \quad k = 3$

Thus, the distance from the vertex to the directrix is:

$\left| \frac{2 \cdot 2 + 1 \cdot 3 - 6}{\sqrt{2^2 + 1^2}} \right| = \left| \frac{4 + 3 - 6}{\sqrt{5}} \right| = \frac{1}{\sqrt{5}}$

The focus of the parabola $P$ is at:

$\left( 2 + \frac{2}{\sqrt{5}}, 3 + \frac{1}{\sqrt{5}} \right)$

Use the eccentricity of the ellipse. The eccentricity $e$ of the ellipse $E$ is given as $\frac{1}{\sqrt{2}}$. For an ellipse,

$e = \frac{\sqrt{a^2 - b^2}}{a}$

Squaring both sides:

$\frac{1}{2} = \frac{a^2 - b^2}{a^2}$ $a^2 - b^2 = \frac{a^2}{2}$ $b^2 = \frac{a^2}{2}$

Calculate the length of the latus rectum. The length of the latus rectum of an ellipse is given by $\frac{2b^2}{a}$. Substituting $b^2 = \frac{a^2}{2}$:

$\text{Latus Rectum} = \frac{2 \cdot \frac{a^2}{2}}{a} = \frac{a^2}{a} = a$

Find $a$ using the focus of the parabola. Since the ellipse passes through the focus of the parabola, substitute the coordinates of the focus into the ellipse equation and solve for $a$ and $b$.

After finding $a$, calculate $\left( \frac{2b^2}{a} \right)^2$ to get the square of the latus rectum.

Thus, the answer is:

$\frac{656}{25}$