A circle of radius 2 unit passes through the vertex and the focus of the parabola $y^{2}=2x$ and touches the parabola $y=\left(x-\frac{1}{4}\right)^{2}+\alpha$, where $\alpha>0$. Then $(4\alpha-8)^{2}$ is equal to ______.

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Parabola

A circle of radius 2 unit passes through the vertex and the focus of the parabola

y^{2} = 2 x

and touches the parabola

y = (x - \frac{1}{4})^{2} + α

, where

α > 0

. Then

(4 α - 8)^{2}

is equal to ______.

Answer: 63

The parabola

y^{2} = 2 x

has vertex

V (0, 0)

and focus

S (\frac{1}{2}, 0)

.

Let the circle be

(x - h)^{2} + (y - k)^{2} = 4

.

Since it passes through

(0, 0)

h^{2} + k^{2} = 4

... (1)

Since it passes through

(\frac{1}{2}, 0)

(\frac{1}{2} - h)^{2} + k^{2} = 4

\Rightarrow h^{2} + k^{2} - h = \frac{15}{4}

... (2)

From (1) and (2):

4 - h = \frac{15}{4} \Rightarrow h = \frac{1}{4}

Then

k^{2} = 4 - \frac{1}{16} = \frac{63}{16} \Rightarrow k = \frac{63}{4}

(taking positive value as the circle must touch the given parabola from above).

The circle touches the parabola

y = (x - \frac{1}{4})^{2} + α

at its vertex

(\frac{1}{4}, α)

.

The distance from center to vertex equals the radius:

α = k + 2 = \frac{63}{4} + 2 = \frac{63 + 8}{4}

Therefore:

4 α - 8 = 63

(4 α - 8)^{2} = 63

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Integer answer