Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$. If the eccentricity of the hyperbola is 5 , then the length of its latus rectum is :

Question

Accepted Answer

A. 1. For the ellipse $E:\dfrac{x^2}{36}+\dfrac{y^2}{16}=1$, here $A^2=36$ and $B^2=16$ (major axis along $x$).

2. For a hyperbola $H:\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ **confocal** with $E$ (sharing the same foci), the relation between their parameters is:
$$
A^2-B^2=a^2+b^2
$$
Substituting:
$$
36-16=a^2+b^2 \ \Rightarrow\ a^2+b^2=20 \quad	ext{...(i)}
$$

3. Given **eccentricity** of the hyperbola is $5$:
$$
e_H^2=1+\frac{b^2}{a^2}=25 \ \Rightarrow\ \frac{b^2}{a^2}=24
$$

4. From (i), $b^2=20-a^2$, so:
$$
\frac{20-a^2}{a^2}=24 \ \Rightarrow\ 20-a^2=24a^2 \ \Rightarrow\ 25a^2=20 \ \Rightarrow\ a^2=\frac45
$$

5. Then:
$$
b^2=20-\frac45=\frac{96}{5}
$$

6. The **latus rectum** is $\dfrac{2b^2}{a}$, where $a=\sqrt{4/5}=\dfrac{2}{\sqrt5}$:
$$
L=\frac{2	imes\frac{96}{5}}{\frac{2}{\sqrt5}}=\frac{96}{5}	imes\sqrt5=\frac{96}{\sqrt5}
$$

Hence, the answer is **Option A**: $\dfrac{96}{\sqrt5}$.

Let the foci of a hyperbola coincide with the foci of the ellipse x236+y216=1\frac{x^2}{36}+\frac{y^2}{16}=136x2​+16y2​=1. If the eccentricity of the hyperbola is 5 , then the length of its latus rectum is :

Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$ . If the eccentricity of the hyperbola is 5 , then the length of its latus rectum is :