The area of the quadrilateral formed with the foci of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ and its conjugate hyperbola is (in square units)

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Aicharya · Accepted Answer

Hyperbola : $\frac{x^2}{16}-\frac{y^2}{9}=1 \Rightarrow e_1=\sqrt{1+\frac{9}{16}}=\frac{5}{4}$ Conjugate hyperbola : $\frac{x^2}{9}-\frac{y^2}{16}=1 \Rightarrow e_2=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$ $\Rightarrow e_2 \gt e_1$ $\Rightarrow$ Area of quadrilateral $=2\left(a^2+b^2\right)=2(16+9)=50$.

The area of the quadrilateral formed with the foci of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ and its…

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