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

Correct Answer: 63

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