Question

lf $\frac{x}{a}+\frac{y}{b}=1$ intersects $5x^2+5y^2+5bx+5ay-9ab=0$ at $P$ and $Q$, $\angle POQ=\frac{\pi }{2}$, then the relation between $a$ and $b$ is

• A. $a=b$
• B. $a=2b$ or $b=2a$
• C. $a=3b$ or $b=3a$
• D. $a+b=5$

Answer 1

The correct option is B $a=2b$ or $b=2a$

Given eq of pair of line

$5x^2+5y^2+5bx+5ay-9ab=0$

and line $\frac{x}{a}+\frac{y}{b}=1$

By homogenisation,

$5x^2+5y^2+5bx(\frac{x}{a}+\frac{y}{b})+5ay(\frac{x}{a}+\frac{y}{b})-9ab(\frac{x}{a}+\frac{y}{b})=0$

$(5+\frac{5b}{a}-\frac{9b}{a})x^2+(5+\frac{5a}{b}-\frac{9a}{b})y^2-8xy=0$

$(5-\frac{4b}{a})x^2+(5-\frac{4a}{b})y^2-8xy=0$

$\therefore \tan \theta =\left|\frac{2\sqrt{4^2-(5-\frac{4b}{a})(5-\frac{4a}{b})}}{10-4(\frac{b}{a}+\frac{a}{b})}\right|$

Given $\theta =90\Rightarrow \tan 90=\infty$

$10=4(\frac{b}{a}+\frac{a}{b})$

$\frac{5}{2}=\frac{a}{b}+\frac{1}{\frac{a}{b}}$

let $t=\frac{a}{b}$

$\frac{5}{2}=t+\frac{1}{t}$

$5t=2(t^2+1)$

$2t^2-5t+2=0$

$t=2,\frac{1}{2}$

$\frac{a}{b}=2$

$a=2b$

or $\frac{a}{b}=\frac{1}{2}$

$2a=b$

$\Rightarrow a=2b$ or $b=2a$