JEE Main 2025 — Hyperbola Question with Solution

Let's find the eccentricities of the given ellipse and hyperbola, and then determine the eccentricity of an ellipse that passes through all four foci.

Step 1: Find $e_1$ for the Ellipse

The equation of the ellipse is:

$\frac{x^2}{b^2}+\frac{y^2}{25}=1$

The eccentricity $e_1$ is given by:

$e_1^2=1-\frac{b^2}{25}$

Step 2: Find $e_2$ for the Hyperbola

The equation of the hyperbola is:

$\frac{x^2}{16}-\frac{y^2}{b^2}=1$

The eccentricity $e_2$ is given by:

$e_2^2=1+\frac{b^2}{16}$

Step 3: Using the Product $e_1{e}_2=1$

Given:

$e_1{e}_2=1$

Thus:

$\left(1-\frac{b^2}{25}\right)\left(1+\frac{b^2}{16}\right)=1$

Expanding gives:

$1+\frac{b^2}{16}-\frac{b^2}{25}-\frac{b^4}{400}=1$

Simplifying:

$\frac{9b^2}{400}=\frac{b^4}{400}$

Thus:

$b^2=9$

Step 4: Determine Eccentricities $e_1$ and $e_2$

Substitute $b^2=9$:

For the ellipse:

$e_1^2=1-\frac{9}{25}=\frac{16}{25}$

$e_1=\frac{4}{5}$

For the hyperbola:

$e_2=\frac{5}{4}$

Step 5: Find the Eccentricity of the New Ellipse

The new ellipse's equation is:

$\frac{x^2}{25}+\frac{y^2}{16}=1$

The eccentricity $e$ is:

$e=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$

Thus, the eccentricity of the ellipse that passes through all four foci is $\frac{3}{5}$.