Question

Show that all chords of the curve $3x^2-y^2-2x+4y=0$, which subtend a right angle at the origin, pass through a fixed point. Find the coordinates of the point.

Answer 1

Let the chord be $y=mx+c$, So find $OA\times OB$ , where $A$ and $B$ are points of intersection of chord and circle , we use homogenisation.

$\Rightarrow 3x^2-y^2-2(x+2y)(y-\frac{mx}{c})=0$

$\Rightarrow 3c{x}^2-c{y}^2-2(xy-m{x}^2+2y^2-2mxy)=0$

$\Rightarrow 3c{x}^2-c{y}^2-2xy-2m{x}^2-4y^2+4mxy=0$

$\Rightarrow x^2(3c-2m)-y^2(c+4)+xy(4m-2)=0$

$\because OA$ and $OB$ areat right angles

$m_1{m}_2=\frac{a}{b}=-1$

$\Rightarrow \frac{3c-2m}{-(c+4)}=-1$

$\Rightarrow 3-2m=c+4$

$\Rightarrow 2c=2(m+2)$

$\Rightarrow c=m+2$

Hence,chord is:

$y=(mx+m+2)$

This equation always satisfies $(-1,2)$ irrespective of value of $m$. Hence the fixed point is $(-1,2)$.