Question

Let each of the two ellipses $E_1:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\;(a>b)$ and $E_2:\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1A$

• A. $\dfrac{16}{5}$
• B. $\dfrac{96}{5}$
• C. $\dfrac{8}{5}$
• D. $\dfrac{32}{5}$

Hint:

Remember that latus rectum depends on the semi-minor axis for major-axis ellipses.

Answer 1

The Correct Option is D

Step 1: Using eccentricity for $E_1$.
\[ e=\frac{c}{a}=\frac{4}{5} \Rightarrow c=\frac{4a}{5} \] Distance between foci is $2c=8$ \[ \Rightarrow c=4 \Rightarrow a=5 \] \[ b^2=a^2-c^2=25-16=9 \] Step 2: Latus rectum of $E_1$.
\[ l_1=\frac{2b^2}{a}=\frac{18}{5} \] Step 3: Using $2l_1^2=9l_2$.
\[ l_2=\frac{2}{9}\left(\frac{18}{5}\right)^2=\frac{72}{25} \] Step 4: Finding focal distance of $E_2$.
\[ l_2=\frac{2A^2}{B},\quad e=\frac{c}{B}=\frac{4}{5} \] Solving gives \[ 2c=\frac{32}{5} \]