Question

A circle of radius 2 unit passes through the vertex and the focus of the parabola y2 = 2x and touches the parabola $y=\left (x−\frac{1}{4}\right)^2+α,$ where $α > 0$. Then $(4α – 8)^2$ is equal to ___________.

Answer 1

The correct answer is 63

Fig.

Assuming the equation of circle is
$x(x-\frac{1}{2})+y^2+λy = 0$
$⇒ x^2+y^2-\frac{1}{2}x+λy = 0$
Radius = \sqrt{\frac{1}{16}+\frac{λ^2}{4}}$$= 2
$⇒ λ^2 = \frac{63}{4}$
$⇒ (x-\frac{1}{4})^2+(y+\frac{λ}{2})^2 = 4$
As this circle and parabola are $y-α = (x-\frac{1}{4})^2$ touching each other.
Hence,
$α = - \frac{λ}{2}+2$
$⇒ (α-2)^2 = \frac{λ^2}{4} = \frac{63}{16}$
$⇒ (4α-8)^2$
= 63