Question

The locus of mid points of the chords of the parabola $y^2=4(x+1)$ which are parallel to $3x=4y$ is

• A. $5y-8=0$
• B. $3x+10=0$
• C. $5y+9=0$
• D. $3y-8=0$

Answer 1

The correct option is D $3y-8=0$

Let the middle points of chord be $(h,k)$
The equation of chord bisected at a point is $T=S_1$
$ky-2(x+h)-4=k^2-4(h+1)$

Slope of chord bisected at the point is $\frac{3}{4}$
$\Rightarrow \frac{2}{k}=\frac{3}{4}$
$\Rightarrow 8-3k=0$
$\therefore$ Locus is $8-3y=0$
$\Rightarrow 3y-8=0$