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

We have,

$y = 4{x}^{2} + 1$

$L: y = x$

Let the foot of perpendicular from $P$ to line $y = x$ is $Q$.

Let $P ≡ \left(x, y\right)$, $Q ≡ \left(c, c\right)$ and $R ≡ \left(h, k\right)$ where, $R$ is the mid-point of $PQ$.

Clearly,

$PQ ⊥ L$

$⇒ \dfrac{k - c}{h - c} = - 1$

$⇒ c = \left(\dfrac{h + k}{2}\right)$

And,

$R ≡ \left(\dfrac{x + c}{2}, \dfrac{y + c}{2}\right)$

$⇒ R ≡ \left(\dfrac{x}{2} + \dfrac{h}{4} + \dfrac{k}{4}, \dfrac{y}{2} + \dfrac{h}{4} + \dfrac{k}{4}\right)$

Hence,

$h = \dfrac{x}{2} + \dfrac{h}{4} + \dfrac{k}{4} ⇒ x = \dfrac{3h}{2} - \dfrac{k}{2}$

$k = \dfrac{y}{2} + \dfrac{h}{4} + \dfrac{k}{4} ⇒ y = \dfrac{3k}{2} - \dfrac{h}{2}$

Now,

$y = 4{x}^{2} + 1$

$⇒ \left(\dfrac{3k - h}{2}\right) = 4{\left(\dfrac{3h - k}{2}\right)}^{2} + 1$

$⇒ \left(3k - h\right) = 2{\left(3h - k\right)}^{2} + 2$

Required locus is

$2{\left(3x - y\right)}^{2} + \left(x - 3y\right) + 2 = 0$