Let P be a variable point on the parabola y = 4x² + 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

Given, parabola $\mathrm{y}=4{\mathrm{x}}^2+1$

Let R (a, b) be mid-point of line joining point P and Q where PQ is perpendicular to line y = x.

Let coordinates of P be P(x, y), Q(q, q) and R(a, b) then,

$\mathrm{a}=\frac{\mathrm{x}+\mathrm{q}}{2}\text{ and }\mathrm{b}=\frac{\mathrm{y}+\mathrm{q}}{2}$

Now, slope of line $\mathrm{y}=\mathrm{x}\text{ is }{\mathrm{m}}_1=1$

Slope of line PQ be

$\frac{\mathrm{b}-\mathrm{q}}{\mathrm{a}-\mathrm{q}}={\mathrm{m}}_2$           (say)

$\because$ Line y = x and PQ are perpendicular to each other,

$\begin{array}{l}\Rightarrow {\mathrm{m}}_1\cdot {\mathrm{m}}_2=-1\\ \Rightarrow \frac{\mathrm{b}-\mathrm{q}}{\mathrm{a}-\mathrm{q}}=-1\Rightarrow \mathrm{b}-\mathrm{q}=\mathrm{q}-\mathrm{a}\\ \Rightarrow \mathrm{q}=\frac{\mathrm{b}+\mathrm{a}}{2}\end{array}$

$\begin{array}{l}\therefore \mathrm{a}=\frac{\mathrm{x}+\mathrm{q}}{2}=\frac{\mathrm{x}+\left(\frac{\mathrm{b}+\mathrm{a}}{2}\right)}{2}=\frac{2\mathrm{x}+\mathrm{b}+\mathrm{a}}{4}\\ \Rightarrow \mathrm{x}=\frac{4\mathrm{a}-\mathrm{b}-\mathrm{a}}{2}=\frac{3\mathrm{a}-\mathrm{b}}{2}\end{array}$

and $\mathrm{b}=\frac{\mathrm{y}+\mathrm{q}}{2}=\frac{\mathrm{y}+\left(\frac{\mathrm{b}+\mathrm{a}}{2}\right)}{2}=\frac{2\mathrm{y}+\mathrm{b}+\mathrm{a}}{4}$

$\Rightarrow \mathrm{y}=\frac{3\mathrm{b}-\mathrm{a}}{2}$

Put (x, y) in equation of parabola as P(x, y) is variable point on parabola

$\begin{array}{l}\frac{3\mathrm{b}-\mathrm{a}}{2}=4{\left(\frac{3\mathrm{a}-\mathrm{b}}{2}\right)}^2+1\\ \frac{(3\mathrm{b}-\mathrm{a})}{2}=(3\mathrm{a}-\mathrm{b}{)}^2+1\\ \Rightarrow (3\mathrm{b}-\mathrm{a})=2(3\mathrm{a}-\mathrm{b}{)}^2+2\end{array}$

Replace $(\mathrm{a}, \mathrm{b}) \mathrm{as} (\mathrm{x},\mathrm{y})\Rightarrow (3\mathrm{y}-\mathrm{x})=2(3\mathrm{x}-\mathrm{y}{)}^2+2$

or   $2(3\mathrm{x}-\mathrm{y}{)}^2+(\mathrm{x}-3\mathrm{y})+2=0$