Question

A circle of radius 2 unit passes through the vertex and the focus of the parabola y² = 2x and touches the parabola y = `(x - 1/4)^2 + α`, where α > 0. Then (4α – 8)² is equal to ______.

Options

• 61

• 62

• 63

• 64

Answer 1

A circle of radius 2 unit passes through the vertex and the focus of the parabola y² = 2x and touches the parabola y = `(x - 1/4)^2 + α`, where α > 0. Then (4α – 8)² is equal to 63.

Explanation:

Given parabola y² = 2x has a vertex V(0, 0) and foci is `S(1/2, 0)`

General equation of circle with a radius of 2 is shown below.

(x – h)² + (y – k)² = 4

∵ Circle passes through (0, 0)

⇒ h² + k² = 4   ...(i)

∵ Circle passes through `(1/2, 0)`, then equation of circle represent as

`(1/2 - h)^2 + k^2` = 4

⇒ h² + k² – h = `15/4`  ...(ii)

On solving (i) and (ii)

4 – h = `15/4` 

⇒ h = `4 - 15/4` = `1/4`

Put the value in equation (i), k = `+ sqrt(63)/4`

k = `- sqrt(63)/4` is rejected as circle with centre `(1/4, - sqrt(63)/4)` can't touch given parabola.

Required equation of circle is `(x - 1/4)^2 + (y - sqrt(63)/4)^2` = 4

When circle touches the parabola y = `(x - 1/4)^2 + α` of vertex `(1/4, α)`

Satisfy `(1/4, α)` in equation of circle

α = `2 + sqrt(63)/4` = `(8 + sqrt(63))/4`

⇒ 4α – 8 = `sqrt(63)`

⇒ (4α – 8)² = 63