Question

All the chords of curve $2x^2+3y^2-5x=0$ which subtend a right angle at the origin are concurrent at:

• A. $(0,1)$
• B. $(1,0)$
• C. $(-1,1)$
• D. $(1,-1)$

Answer 1

The correct option is B $(1,0)$

The equation of the curve is $2x^2+3y^2-5x=0$
Let the equation of the chords of curve be $y=mx+c$
To homogenize,
$2x^2+3y^2-5x\left(\frac{y-mx}{c}\right)=0\Rightarrow (2c+5m)x^2+\left(3c\right)y^2-5xy=0$
These lines are perpendicular hence
$(2c+5m)+\left(3c\right)=0\Rightarrow 5c+5m=0\Rightarrow c=-m$
Substituting this in equation of chord, we get
$y=mx-m\Rightarrow y=m(x-1)$
which is of the form $P+\lambda Q=0$
For all real values of $m$ the chord passes through the point of intersection of lines
$y=0$ and $x-1=0$
Solving these we get point of intersection as $(1,0)$