Let P be a variable point on the parabola y=4x^2+1.

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

Accepted Answer

Let $P=(t,4t^2+1)$. Foot of perpendicular to $y=x$: $\left(\dfrac{t+4t^2+1}2,\dfrac{t+4t^2+1}2\right)=(s,s)$.

Midpoint $M=(X,Y)$ of $P$ and $(s,s)$: $X=\dfrac{4t^2+3t+1}4$, $Y=\dfrac{12t^2+t+3}4$.

$$3X-Y=2t,\qquad X-3Y=-8t^2-2.$$

Substituting $t=\dfrac{3X-Y}2$: $8\left(\dfrac{3X-Y}2\right)^2=-(X-3Y)-2$, i.e.
$2(3X-Y)^2+(X-3Y)+2=0.$

Solutions · 1