JEE Mathematics 2024 Question with Solution

Given: The ellipse is $\frac{(x-1)^2}{100} + \frac{(y-1)^2}{75} = 1$. The hyperbola $H$ has the same foci as the ellipse, and its eccentricity is the reciprocal of the ellipse's eccentricity.

Find: The value of $3\alpha^2 + 2\beta^2$, where $\alpha$ is the transverse axis length and $\beta$ is the conjugate axis length of $H$.

For the ellipse,

Since the larger denominator is under $x$, the major axis is along the $x$-axis and the center is $(1,1)$.

Its eccentricity is

Hence the ellipse has

So the foci are $(1 \pm 5, 1)$, that is, $(-4,1)$ and $(6,1)$.

For the hyperbola, the eccentricity is the reciprocal of $\frac{1}{2}$, so

The hyperbola has the same foci, hence

Using $e_H = \frac{c}{a}$,

Now use the hyperbola relation

So,

Therefore, the transverse axis length is

and the conjugate axis length is

Now,

Hence,

Therefore, the correct option is B.

Given: The ellipse has $a^2 = 100$ and $b^2 = 75$.

Find: $3\alpha^2 + 2\beta^2$ for the hyperbola with the same foci and reciprocal eccentricity.

From the ellipse,

So for the hyperbola,

Also, the common focal distance is

Since for the hyperbola $e_2 = \frac{c}{a}$,

Thus,

Now use

So,

With $a = \frac{5}{2}$,

Hence,

Therefore,

So the correct option is B.