Question

Let $P$ be a variable point on the parabola $y=4x^2+1.$ 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. ${(3x-y)}^2+2(x-3y)+2=0$
• B. $2{(3x-y)}^2+(x-3y)+2=0$
• C. $2{(x-3y)}^2+(3x-y)+2=0$
• D. ${(3x-y)}^2+(x-3y)+2=0$

Answer 1

The correct option is B $2{(3x-y)}^2+(x-3y)+2=0$


Let coordinate of mid-point $M$ of $PQ$ be $(x_1,y_1)$ Let coordinate of $Q$ be $(\alpha ,\beta ).$
$\therefore \frac{\alpha -x_1}{1}=\frac{\beta -y_1}{-1}=\frac{-(x_1-y_1)}{2}$
$\therefore Q=(\alpha ,\beta )=(\frac{x_1+y_1}{2},\frac{x_1+y_1}{2})$
and coordinate of $P=(\frac{3x_1-y_1}{2},\frac{3y_1-x_1}{2})$
$\therefore$ $P$ lies on parabola $y=4x^2+1$
$\therefore \frac{3y_1-x_1}{2}=4{\left(\frac{3x_1-y_1}{2}\right)}^2+1$
$\therefore$ Required locus is $2{(3x-y)}^2+(x-3y)+2=0$