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Circle is drawn with end points of latus...

Circle is drawn with end points of latus rectum of the parabola `y^2 = 4ax` as diameter, then equation of the common tangent to this circle & the parabola `y^2 = 4ax` is :

A

`y = x + a`

B

`x - sqrt(3)y + 3a = 0`

C

`4x - 2y + a = 0`

D

`3x - sqrt(3)y + a = 0`

Text Solution

Verified by Experts

The correct Answer is:
B
Equation of circle is `(x - a) (x -a ) + (y - 2a) (y + 2a) = 0`
or `x^2 + y^2 - 2ax - 3a^2 = 0`.
`y = mx + a/m` be the common tangent then its distance from (a, 0) must be 2a.
`|(ma + a/m)/(sqrt(1 + m^2))| = 2a`
`|(sqrt(m^2 + 1))/(m)| = 2 implies m = +- 1/(sqrt3)`
`y = 1/(sqrt(3))x + sqrt(3) a or y =(-1)/(sqrt(3)) x - sqrt(3)a`
or, `x - sqrt(3)y + 3a =0 or x + sqrt(3) y + 3a = 0`
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