Question

If the line $\frac{x}{a}+\frac{y}{b}=1$ intersects the curve
$5x^2+5y^2+5bx+5ay-9ab=0$ at $P$ and $Q$ such that $\angle POQ={90}^{\circ },$ where $O$ is the origin, then the value of $\frac{a}{b}$ is

• A. $\frac{1}{2}$
• B. $2$
• C. $\frac{1}{3}$
• D. $3$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{1}{2}$
B $2$

$\frac{x}{a}+\frac{y}{b}=1\cdots \left(1\right)$
$5x^2+5y^2+5(bx+ay)1-9ab(1{)}^2=0\cdots (2)$
Combined equation of lines $OP$ and $OQ$ can be obtained by homogenization of $\left(2\right)$ w.r.t. to $\left(1\right)$
$5x^2+5y^2+5(bx+ay)(\frac{x}{a}+\frac{y}{b})-9ab{(\frac{x}{a}+\frac{y}{b})}^2=0$
$\Rightarrow (5-\frac{4b}{a})x^2+(5-\frac{4a}{b})y^2-8xy=0$
Since, lines are perpendicular,
$\Rightarrow 10=4(\frac{a}{b}+\frac{b}{a})$
$\Rightarrow \frac{a}{b}+\frac{b}{a}=\frac{5}{2}$

Let $\frac{a}{b}=t$
$\Rightarrow t+\frac{1}{t}=\frac{5}{2}$
$\Rightarrow 2t^2-5t+2=0$
$\Rightarrow (2t-1)(t-2)=0$
$\Rightarrow t=\frac{a}{b}=\frac{1}{2}$ or $2$