Let P be a parabola with vertex (2, 3) and directrix 2x + y = 6. Let an ellipse , a > b of eccentricity pass through the focus of the parabola P. Then the square of the length of the latus rectum of E is [2024]

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Solution

Q.
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 [2024]

1 $\frac{385}{8}$

2 $\frac{656}{25}$

3 $\frac{347}{8}$

4 $\frac{512}{25}$

Ans.
(2)

Given that vertex of parabola P is (2, 3) and equation of directrix is 2x + y = 6 ... (i)

Let S( $α$ , $β$ ) be the focus of parabola.

Equation of axis of AS is

$y - 3 = \frac{1}{2} (x - 2)$

$\Rightarrow 2 y - 6 = x - 2$

$\Rightarrow x - 2 y + 4 = 0$ ... (ii)

Solving equations (i) and (ii), we get

$∴ x = \frac{8}{5} and y = \frac{14}{5} ∴ A \equiv (\frac{8}{5}, \frac{14}{5})$

Now, using mid point formula, we have

$\frac{\frac{8}{5} + α}{2} = 2; \frac{\frac{14}{5} + β}{2} = 3 \Rightarrow \frac{8}{5} + α = 4; \frac{14}{5} + β = 6$

$\Rightarrow α = 4 - \frac{8}{5} and β = 6 - \frac{14}{5} \Rightarrow α = \frac{12}{5}; and β = \frac{16}{5}$

The point $S (\frac{12}{5}, \frac{16}{5})$ will satisfy the ellipse.

$∴ \frac{144}{25 a^{2}} + \frac{256}{25 b^{2}} = 1$ ... (iii)

Now, $b^{2} = a^{2} (1 - e^{2})$

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

$∴$ From (iii), $\frac{144}{25 \times 2 b^{2}} + \frac{256}{25 b^{2}} = 1$

$\Rightarrow \frac{72}{25} + \frac{256}{25} = b^{2} \Rightarrow b^{2} = \frac{328}{25} ∴ a^{2} = \frac{2 \times 328}{25} = \frac{656}{25}$

$∴$ Length of latus rectum $= \frac{2 b^{2}}{a} = \frac{2 \times 328}{25 \times \frac{\sqrt{656}}{5}} = \frac{656 \times 5}{25 \times \sqrt{656}} = \frac{\sqrt{656}}{5}$

$∴$ Square of the latus rectum $= \frac{656}{25}$ .

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