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# Let A be the vertex and L the length of the latus rectum of the parabola y^{2} – 2y – 4x – 7 = 0. The equation of the parabola with A as vertex, 2L the length of the latus rectum and the axis at right angles to that of the given curve is -

- x
^{2} + 4x + 8y – 4 = 0
- x
^{2} + 4x + 8y – 12 = 0
- x
^{2} + 4x + 8y + 12 = 0
- x
^{2} + 8x – 4y + 8 = 0

^{2}+ 4x + 8y – 4 = 0^{2}+ 4x + 8y – 12 = 0^{2}+ 4x + 8y + 12 = 0^{2}+ 8x – 4y + 8 = 0Hint:

### Let (y-k)^{2} = 4a(x-h) the (h,k) is the vertex and 4a is the length of latus rectum of the parabola and y = k is the axis of the parabola.

## The correct answer is: x^{2} + 4x + 8y – 4 = 0

### Given, the equation y^{2} – 2y – 4x – 7 = 0

y^{2 }- 2y +1 = 4x + 7+ 1

(y-1)^{2} = 4(x+2)

here length of latus rectum L = 4 , vertex of parabola = (-2,1) and axis of parabola is y = 1

Then equation for perpendicular y = 1 and veterx (-2,1) is x = -2

parabola length of latus rectum = 2L = 2*4 = 8

the equation of parabola with axis x = -2 and latus rectum = 8 and vertex (-2,1) is

Therefore, x^{2} + 4x + 8y – 4 = 0 is the parabola that with stand the coditions.

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