Question

Let $P$ be the foot of the perpendicular from focus $S$ of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ on the line $bx-ay=0$ and let $C$ be the centre of the hyperbola. Then the area of the rectangle whose sides are equal to that of $SP$ and $CP$ is

• A. $2ab$
• B. $ab$
• C. $\frac{a^2+b^2}{2}$
• D. $\frac{a}{b}$

Answer 1

The correct option is B $ab$

$P$ be the foot of the perpendicular from focus $S$ of hyperbola
$\therefore SP=\frac{b\cdot a\cdot e-0}{\sqrt{b^2+a^2}}=\frac{a\cdot b\cdot e}{\sqrt{a^2(e^2-1)+a^2}}=b$
$C$ be the centre of the hyperbola
$\therefore CP=\sqrt{a^2{e}^2-b^2}=\sqrt{a^2{e}^2-a^2(e^2-1)}=a$
Area of the rectangle whose sides are equal to that of $SP$ and $CP=a\cdot b$