Question ID: #856
Let each of the two ellipses $E_{1}:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,(a>b)$ and $E_{2}:\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1 (A \lt B)$ have eccentricity $\frac{4}{5}$. Let the lengths of the latus recta of $E_{1}$ and $E_{2}$ be $l_{1}$ and $l_{2},$ respectively, such that $2l_{1}^{2}=9l_{2}$. If the distance between the foci of $E_{1}$ is 8, then the distance between the foci of $E_{2}$ is
- (1) $\frac{96}{5}$
- (2) $\frac{32}{5}$
- (3) $\frac{16}{5}$
- (4) $\frac{8}{5}$
Solution:
For Ellipse $E_1$ ($a>b$):
Eccentricity $e = \frac{4}{5}$.
Distance between foci $= 2ae = 8 \Rightarrow ae = 4 \Rightarrow a(4/5) = 4 \Rightarrow a = 5$.
Using $b^2 = a^2(1-e^2)$:
$$b^2 = 25\left(1 – \frac{16}{25}\right) = 25\left(\frac{9}{25}\right) = 9 \Rightarrow b=3$$
Length of latus rectum $l_1 = \frac{2b^2}{a} = \frac{2(9)}{5} = \frac{18}{5}$.
For Ellipse $E_2$ ($A
$$2\left(\frac{18}{5}\right)^2 = 9l_2$$
$$2 \left(\frac{324}{25}\right) = 9l_2$$
$$l_2 = \frac{2 \times 324}{9 \times 25} = \frac{2 \times 36}{25} = \frac{72}{25}$$
Equating the expressions for $l_2$:
$$\frac{18B}{25} = \frac{72}{25}$$
$$18B = 72 \Rightarrow B = 4$$
The distance between the foci of $E_2$ is $2Be$.
$$2Be = 2(4)\left(\frac{4}{5}\right) = \frac{32}{5}$$
Ans. (2)
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