JEE Mathematics 2025 Question with Solution

Given: The ellipse is $\frac{x^2}{b^2} + \frac{y^2}{25} = 1$ and the hyperbola is $\frac{x^2}{16} - \frac{y^2}{b^2} = 1$ with $b < 5$ and $e_1 e_2 = 1$.

Find: The eccentricity of the ellipse with axes along the coordinate axes that passes through the four foci of the given conics.

For the ellipse,

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

For the hyperbola,

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

Using $e_1 e_2 = 1$,

$$\sqrt{1 - \frac{b^2}{25}} \cdot \sqrt{1 + \frac{b^2}{16}} = 1$$

Squaring both sides,

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

Expanding,

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

So,

$$\frac{b^2}{16} - \frac{b^2}{25} = \frac{b^4}{400}$$ $$b^2\left(\frac{1}{16} - \frac{1}{25}\right) = \frac{b^4}{400}$$ $$\frac{9b^2}{400} = \frac{b^4}{400}$$

Hence,

$$b^2 = 9$$

Since $b < 5$,

$$b = 3$$

Now the first ellipse becomes $\frac{x^2}{9} + \frac{y^2}{25} = 1$, so its foci are

$$(0, \pm 4)$$

The hyperbola becomes $\frac{x^2}{16} - \frac{y^2}{9} = 1$, so its foci are

$$(\pm 5, 0)$$

Therefore the required ellipse, with axes along the coordinate axes and passing through all four points, is

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

Its eccentricity is

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

Therefore, the eccentricity of the required ellipse is $\frac{3}{5}$. The correct option is B.

Given: $e_1 = \sqrt{1 - \frac{b^2}{25}}$ for the ellipse and $e_2 = \sqrt{1 + \frac{b^2}{16}}$ for the hyperbola, with $e_1 e_2 = 1$.

Find: The eccentricity of the new ellipse through all four foci.

From the condition,

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

This gives

$$1 - \frac{b^2}{25} + \frac{b^2}{16} - \frac{b^4}{400} = 1$$ $$\frac{25b^2 - 16b^2}{400} = \frac{b^4}{400}$$ $$\frac{9b^2}{400} = \frac{b^4}{400}$$

So,

$$b^2 = 9$$

and hence $b = 3$.

Now for $\frac{x^2}{9} + \frac{y^2}{25} = 1$, the focal distance is

$$c = \sqrt{25 - 9} = 4$$

so the foci are $(0, \pm 4)$.

For $\frac{x^2}{16} - \frac{y^2}{9} = 1$, the focal distance is

$$c = \sqrt{16 + 9} = 5$$

so the foci are $(\pm 5, 0)$.

An ellipse with axes along the coordinate axes passing through $(\pm 5, 0)$ and $(0, \pm 4)$ must be

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

Therefore its eccentricity is

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

Therefore, the required eccentricity is $\frac{3}{5}$.