JEE Mathematics 2024 Question with Solution

Given: Hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ and ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The ellipse passes through the foci of the hyperbola and satisfies $e_1e_2 = 1$.

Find: The length of the chord of the ellipse parallel to the $x$-axis through $\left(0,2\right)$.

For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we have $a^2 = 16$ and $b^2 = 9$. Hence,

The foci of the hyperbola are at $\left(\pm ae_1,0\right)$. Since $a = 4$,

So the foci are $\left(\pm 5,0\right)$.

The ellipse passes through $\left(5,0\right)$. Substituting into $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,

Therefore,

Given $e_1e_2 = 1$,

so,

For the ellipse,

Substituting $e_2 = \frac{4}{5}$ and $a^2 = 25$,

Hence,

So the ellipse is

For the chord parallel to the $x$-axis through $\left(0,2\right)$, put $y = 2$ in the ellipse:

Thus,

The chord length is $2x$, so

Therefore, the correct option is C and the length of the chord is $\frac{10\sqrt{5}}{3}$.

Given: The ellipse passes through the foci $\left(\pm 5,0\right)$ of the hyperbola and satisfies $e_1e_2 = 1$.

Find: The chord length parallel to the $x$-axis through $\left(0,2\right)$.

From the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$,

Hence,

Since the ellipse passes through $\left(\pm 5,0\right)$, its semi-major axis is $a = 5$. Using

we get $b = 3$. Therefore the ellipse is $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

Now use the chord formula for the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at ordinate $y$:

At $y = 2$,

Therefore, the correct option is C.