Given ellipse

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

Comparing with standard form,

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

we get

$a^2=25,\quad b^2=16 \Rightarrow a=5,\ b=4$

Find the foci of the ellipse

For the ellipse,

$c^2=a^2-b^2=25-16=9 \Rightarrow c=3$

Hence the foci are

$(\pm 3,0)$

Form of the required hyperbola

Its transverse and conjugate axes coincide with the major and minor axes of the ellipse respectively, so the hyperbola has transverse axis along the $x$-axis and conjugate axis along the $y$-axis.

Therefore its equation is of the form

$\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$

Hyperbola passes through the foci of the ellipse

Since $(\pm 3,0)$ lie on the hyperbola, substitute $(3,0)$:

$\frac{3^2}{A^2}-0=1$

$\frac{9}{A^2}=1 \Rightarrow A^2=9$

So the hyperbola is

$\frac{x^2}{9}-\frac{y^2}{B^2}=1$

Use the condition on eccentricities

• Eccentricity of the ellipse:

  $e_1=\frac{c}{a}=\frac{3}{5}$

• Let eccentricity of the hyperbola be $e_2$.

  Given product of eccentricities is $1$:

  $e_1 e_2=1$

  $\frac{3}{5}e_2=1 \Rightarrow e_2=\frac{5}{3}$

Relate eccentricity of hyperbola to its parameters

For hyperbola,

$e_2=\sqrt{1+\frac{B^2}{A^2}}$

With $A^2=9$,

$\sqrt{1+\frac{B^2}{9}}=\frac{5}{3}$

Squaring,

$1+\frac{B^2}{9}=\frac{25}{9}$

$\frac{B^2}{9}=\frac{16}{9} \Rightarrow B^2=16$

Equation of the hyperbola

$\boxed{\frac{x^2}{9}-\frac{y^2}{16}=1}$

Check options

This matches Option B.

Comparison with stored answer

Stored correct answer: B

Derived answer: B

So they agree.