Question

Consider an ellipse \[ E_1:\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \ (a>b) \quad \text{and} \quad E_2:\ \frac{x^2}{A^2}+\frac{y^2}{B^2}=1 \ (B>A), \] where $e=\dfrac{4}{5}$ for both the curves and $\ell_1$ is the length of latus rectum of $E_1$ and $\ell_2$ is the length of latus rectum of $E_2$. Let the distance between the foci of the first curve be $8$. Find the distance between the foci of the second curve. (Given $2\ell_1^2=9\ell_2$).

• A. $\dfrac{64}{5}$
• B. $\dfrac{8}{5}$
• C. $\dfrac{32}{5}$
• D. $\dfrac{16}{5}$

Hint:

For an ellipse, always remember: eccentricity relates semi-major axis and focal distance, while latus rectum links both axes directly.

Answer 1

The Correct Option is C

Step 1: Use eccentricity for the first ellipse.
For $E_1$, eccentricity \[ e=\frac{c_1}{a}=\frac{4}{5} \Rightarrow c_1=\frac{4a}{5} \] Distance between foci is $2c_1=8$: \[ 2\cdot\frac{4a}{5}=8 \Rightarrow a=5 \] Step 2: Find $b$ for the first ellipse.
\[ b^2=a^2(1-e^2)=25\left(1-\frac{16}{25}\right)=9 \] Step 3: Find latus rectum of $E_1$.
\[ \ell_1=\frac{2b^2}{a}=\frac{2\times9}{5}=\frac{18}{5} \] Step 4: Use the given relation to find $\ell_2$.
\[ 2\ell_1^2=9\ell_2 \Rightarrow 2\left(\frac{18}{5}\right)^2=9\ell_2 \Rightarrow \ell_2=\frac{72}{25} \] Step 5: Find parameters of the second ellipse.
For $E_2$, $e=\dfrac{c_2}{B}=\dfrac{4}{5}$ and \[ \ell_2=\frac{2A^2}{B}=\frac{72}{25} \] Also, \[ A^2=B^2(1-e^2)=B^2\left(1-\frac{16}{25}\right)=\frac{9B^2}{25} \] Substitute: \[ \frac{2}{B}\cdot\frac{9B^2}{25}=\frac{72}{25} \Rightarrow B=4 \] Step 6: Find the distance between the foci of $E_2$.
\[ c_2=\frac{4B}{5}=\frac{16}{5} \Rightarrow \text{distance}=2c_2=\frac{32}{5} \]