Question:
Let each of the two ellipses $E_1:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\;(a>b)$ and $E_2:\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1A$
Let each of the two ellipses $E_1:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\;(a>b)$ and $E_2:\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1A$
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Remember that latus rectum depends on the semi-minor axis for major-axis ellipses.
Updated On: Mar 5, 2026
- $\dfrac{16}{5}$
- $\dfrac{96}{5}$
- $\dfrac{8}{5}$
$\dfrac{32}{5}$
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The Correct Option is D
Solution and Explanation
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 \[ 2c=\frac{32}{5} \]
\[ e=\frac{c}{a}=\frac{4}{5} \Rightarrow c=\frac{4a}{5} \] 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 \[ 2c=\frac{32}{5} \]
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