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,
$SP=\left|\frac{abe}{\sqrt{b^2+a^2}}\right|$
$=\left|\frac{abe}{ae}\right|=b$
and $CS=ae$
Again, $△SPC$ is right angled triangle at $P$.
$\therefore CP=\sqrt{C{S}^2-S{P}^2}$
$=\sqrt{a^2{e}^2-b^2}$
$=\sqrt{a^2\left(1+\frac{b^2}{a^2}\right)-b^2}$
$=\sqrt{a^2+b^2-b^2}=a$
$\therefore$ Area of rectangle $=CP\times SP$
$=ab$