about to only mathematics

Options

A. `x^(2) + 2sqrt3y = pm 3sqrt3`
B. `x^(2) pm 2sqrt3y = 3 pm sqrt3`
C. `x^(2) + 2 sqrt3y = sqrt3 pm 3`
D. `x^(2) - 2sqrt3y = pm 3sqrt3`

Answer

B

Explanation

We have,
`x^(2) + 4y^(2) = 4 rArr x^(2)/2^(2) + y^(2)/1^(2) = 1`
Let e be the eccentricity of this ellispe. Then,
`e^(2) = 1 - 1/4 = 3/4 rArr e = sqrt3/2`
Clearly, coordinates of P and Q are `(sqrt3, -1//2)` and `(-sqrt3, -1//2)` respectively.

`therefore PQ = 2sqrt3`
`rArr` Latusrectum of the parabola `= 2sqrt3`
Cooridnates of foci of two parabolas are `(0, -1//2)` and `(0, -3//2)`.
So, the coordinates of the vertices of two parabola are `(0, (-1 - sqrt3)/(2)) and (0 (-1 + sqrt(3))/(2)).`
Hence, their equations are
`(x - 0)^(2) = 2sqrt3 (y + (1 + sqrt3)/(2)) and (x - 0)^(2) = -2sqrt(3)(y + (1 - sqrt3)/(2))`
or `x^(2) = 2sqrt3y = 3 + sqrt3` and `x^(2) + 2sqrt3 y = 3 - sqrt3`