Question

if all the chord of the curve $3x^2-y^2-2x+4y=0$ which subtend a right angle at origin passes through fixed point $(a,b)$. Then $|a+b|$ is equal to:

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

Given curve $3x^2-y^2-2x+4y=0\cdots \left(i\right)$
Let the chord equation of the curve be $lx+my=1\cdots \left(ii\right)$
Homogenising $\left(i\right)$ with the help of $\left(ii\right)$
$3x^2-y^2-(2x-4y)(lx+my)=0$
$\Rightarrow (3-2l)x^2+(-1+4m)y^2+(-2m+4l)xy=0$
$\because$ it subtends right angle at the origin
$\Rightarrow (3-2l)+(-1+4m)=0$
$\Rightarrow 2m-l+1=0$
$\Rightarrow l-2m=1\cdots \left(iii\right)$
from $\left(ii\right)$ and $\left(iii\right)$, chord always passes through $(1,-2)$
$\Rightarrow |a+b|=|-1|=1$