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 hyperbola. Then the area of the rectangle whose sides are equal to that of $SP$ and $CP$ is

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

Answer 1

Answer: (B)

Given, equation of hyperbola is
$\frac{x^2}{a62} - \frac{y^2}{b^2} = 1$

From figure,
$S P =\left|\frac{a b e}{\sqrt{b^{2}+a^{2}}}\right| $
$=\left|\frac{a b e}{a e}\right|=b$
and $C S=a e$
Again, $\triangle S P C$ is right angled triangle at $P$.
$\therefore C P =\sqrt{C S^{2}-S P^{2}} $
$= \sqrt{a^{2} e^{2}-b^{2}} $
$= \sqrt{a^{2}\left(1+\frac{b^{2}}{a^{2}}\right)-b^{2}}$
$=\sqrt{a^{2}+b^{2}-b^{2}}=a $
$ \therefore $ Area of rectangle $=C P \times S P $
$=a b $