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(A) Given that $p$ and $q$ are the eccentricities of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate hyperbola respectively.We kno

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(A) Given that $p$ and $q$ are the eccentricities of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate hyperbola respectively.We know that $p^2 = 1 + \frac{b^2}{a^2} = \frac{a^2+b^2}{a^2}$ and $q^2 = 1 + \frac{a^2}{b^2} = \frac{a^2+b^2}{b^2}$.The equation of the ellipse is $\frac{x^2}{p^2} + \frac{y^2}{q^2} = 1$.Substituting the values of $p^2$ and $q^2$,we get $\frac{x^2 a^2}{a^2+b^2} + \frac{y^2 b^2}{a^2+b^2} = 1$,which simplifies to $a^2 x^2 + b^2 y^2 = a^2 + b^2$.Given the pair of lines $x^2 - y^2 = 0$,which implies $y^2 = x^2$.Substituting $y^2 = x^2$ into the ellipse equation: $a^2 x^2 + b^2 x^2 = a^2 + b^2$.$(a^2 + b^2) x^2 = a^2 + b^2 \implies x^2 = 1 \implies x = \pm 1$.Since $y^2 = x^2$,we have $y = \pm 1$.The points of intersection are $(1, 1), (1, -1), (-1, 1), (-1, -1)$.These points form a square with side length $s = \sqrt{(1 - (-1))^2 + (1 - 1)^2} = 2$.Area of the square = $s^2 = 2^2 = 4$ sq. units.

If $p$ and $q$ are the eccentricities of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate hyperbola respectively,then the area of the square (in sq. units) formed by the points of intersection of the ellipse $\frac{x^2}{p^2}+\frac{y^2}{q^2}=1$ and the pair of lines $x^2-y^2=0$ is

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