Question

All chords of the curve $3x^2-y^2-2x+4y=0$ which subtend a right angle at the origin, pass through the fixed point $(a,b)$ which is equivalent to

• A. $(1,2)$
• B. $(1,-2)$
• C. $(2,1)$
• D. $(-2,1)$

Answer 1

Answer: (B)

$(1,-2)$

Let $y=mn+c$ is chord of $3x^2-y^2-2x+4y=0$

$\therefore \frac{y-mn}{c}=1$

By homogenisation,

$3x^2-y^2-2x\left(\frac{y-mn}{c}\right)+4y\left(\frac{y-mn}{c}\right)=0$

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

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

Given $\theta ={90}^{\circ }$

$\tan \theta =0$
$\therefore a+b=0$

$(3c+2m)+(4-c)=0$

$2c+2m+4=0$

$-2=m+c$
$\therefore y=-2$ and $x=1$

Hence, option B is the correct answer.