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A parabola is drawn through two given po...

A parabola is drawn through two given points `A(1,0)` and `B(-1,0)` such that its directrix always touches the circle `x² + y^2 = 4.` Then, the maximum possible length of semi latus rectum is

A

`2+sqrt(3)`

B

`3+sqrt(3)`

C

`4+sqrt(3)`

D

`1+sqrt(3)`

Text Solution

Verified by Experts

The correct Answer is:
A
Any point on circle `x^(2) + y^(2) =4` is `(2 cos alpha, 2 sin alpha)` So, equation of directrix is `x (cos alpha) + y (sin alpha) -2 =0`.
Let focus be `(x_(1),y_(1))`.
Then as `A(1,0), B(-1,0)` lie on parabola, we must have
`(x_(1)-1)^(2) + y_(1)^(2) = (cos alpha -2)^(2)` (using definition of parabola) and `(x_(1)+1)^(2) + y_(1)^(2) = (cos alpha +2)^(2)`
`rArr x_(1) = 2 cos alpha, y_(1) = +- sqrt(3) sin alpha`
Thus, locus of focus is `(x^(2))/(4) + (y^(2))/(3) =1` and focus is of the form `(2 cos alpha, +- sqrt(3) sin alpha)`.
`:.` Length of semi latus rectum of parabola = perpendicular distance from focus to directrix `=|2 +-sqrt(3)| sin^(2) alpha`
Hence maximum possible length `= 2+ sqrt(3)`
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