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. $\frac{(a^2+b^2)}{2}$
• D. $\frac{a}{b}$

Answer 1

The correct option is B $ab$

The centre of the hyperbola is $C(0,0)$. Let the coordinates of the focus $S$ be $(s,0)$ and assume that $a>b$.

$\therefore s=\sqrt{a^2+b^2}$

Also, perpendicular from focus to asymptote $bx-ay=0⟹y=\frac{b}{a}x$ has slope $-\frac{a}{b}$, we have equation of normal as

$y=-\frac{a}{b}x+k$

$⟹by+ax=bk=a\sqrt{a^2+b^2}$ (as passes through the focus)

Hence, coordinates of $P$ are $(\frac{a^2}{\sqrt{a^2+b^2}},\frac{ab}{\sqrt{a^2+b^2}})$

$\therefore SP=\sqrt{{(\sqrt{a^2+b^2}-\frac{a^2}{\sqrt{a^2+b^2}})}^2+{(0-\frac{ab}{\sqrt{a^2+b^2}})}^2}$

$⟹SP=\sqrt{\frac{b^4}{a^2+b^2}+\frac{a^2{b}^2}{a^2+b^2}}$

$⟹SP=\sqrt{b^2\left(\frac{b^2+a^2}{a^2+b^2}\right)}$

$⟹SP=b$

Also,

$CP=\sqrt{{(0-\frac{a^2}{\sqrt{a^2+b^2}})}^2+{(0-\frac{ab}{\sqrt{a^2+b^2}})}^2}$

$⟹CP=\sqrt{\frac{a^4}{a^2+b^2}+\frac{a^2{b}^2}{a^2+b^2}}$

$⟹CP=\sqrt{a^2\left(\frac{a^2+b^2}{a^2+b^2}\right)}$

$⟹CP=a$

Hence, area of rectangle with sides $CP$ and $SP$ is $ab$.