# Let P be a variable point

Question:

Let $\mathrm{P}$ 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 $\mathrm{P}$ to the line $\mathrm{y}=\mathrm{x}$ is :

1. $(3 x-y)^{2}+(x-3 y)+2=0$

2. $2(3 x-y)^{2}+(x-3 y)+2=0$

3. $(3 x-y)^{2}+2(x-3 y)+2=0$

4. $2(x-3 y)^{2}+(3 x-y)+2=0$

Correct Option: , 2

Solution:

$\frac{\mathrm{K}-\mathrm{C}}{\mathrm{h}-\mathrm{C}}=-1$

$\mathrm{C}=\frac{\mathrm{h}+\mathrm{K}}{2} \quad \mathrm{P}(\mathrm{x}, \mathrm{y})$

$R=\left(\frac{x+C}{2}, \frac{y+C}{2}\right)$

$R=\left(\frac{x}{2}+\frac{h}{4}+\frac{K}{4}, \frac{y}{2}+\frac{h}{4}+\frac{k}{4}\right)$

$h=\frac{x}{2}+\frac{h}{4}+\frac{K}{4}$

$\mathrm{K}=\frac{\mathrm{y}}{2}+\frac{\mathrm{h}}{4}+\frac{\mathrm{K}}{4}$

$\Rightarrow x=\frac{3 h}{2}-\frac{K}{2}, y=\frac{3 K}{2}-\frac{h}{2}$

$\mathrm{Y}=4 \mathrm{x}^{2}+1$

$\left(\frac{3 \mathrm{k}-\mathrm{h}}{2}\right)=4\left(\frac{3 \mathrm{~h}-\mathrm{k}}{2}\right)^{2}+1$