Question

Let $P$ be a variable point on the parabola $y=4 x^{2}+1$. Then, the locus of the mid-point of the point $P$ and the foot of the perpendicular drawn from the point $P$ to the line $y=x$ is :

• A. $(3 x-y)^{2}+(x-3 y)+2=0$
• B. $2(3 x-y)^{2}+(x-3 y)+2=0$
• C. $(3 x-y)^{2}+2(x-3 y)+2=0$
• D. $2(x-3 y)^{2}+(3 x-y)+2=0$

Answer 1

Answer: (B)

$\frac{K-C}{h-C}=-1$
$C=\frac{h+K}{2}$
$R=\left(\frac{x+C}{2}, \frac{y+C}{2}\right)$
$R=\left(\frac{x}{2}+\frac{h}{4}+\frac{K}{4}, \frac{y}{2}+\frac{h}{4}+\frac{k}{4}\right)$
$h=\frac{x}{2}+\frac{h}{4}+\frac{K}{4}$
$K=\frac{y}{2}+\frac{h}{4}+\frac{K}{4}$
$\Rightarrow x=\frac{3 h}{2}-\frac{K}{2}, y=\frac{3 K}{2}-\frac{h}{2}$
$Y=4 x^{2}+1$
$\left(\frac{3 k-h}{2}\right)=4\left(\frac{3 h-k}{2}\right)^{2}+1$