Let A be the vertex and L be the length of th | vedclass

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(D) Given parabola: $y^2 - 2y - 4x - 7 = 0$$(y - 1)^2 = 4x + 8 = 4(x + 2)$Vertex $A = (-2, 1)$ and length of latus rectum $L = 4$.For the new parabola

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Both $(A)$ and $(B)$

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(D) Given parabola: $y^2 - 2y - 4x - 7 = 0$$(y - 1)^2 = 4x + 8 = 4(x + 2)$Vertex $A = (-2, 1)$ and length of latus rectum $L = 4$.For the new parabola,the vertex is $A(-2, 1)$,the length of latus rectum is $2L = 8$,and the axis is perpendicular to the axis of the given parabola $(y = 1)$.Thus,the axis of the new parabola is $x = -2$.The equation of the new parabola is $(x + 2)^2 = \pm 4(2)(y - 1)$.Case $1$: $(x + 2)^2 = 8(y - 1) \implies x^2 + 4x + 4 = 8y - 8 \implies x^2 + 4x - 8y + 12 = 0$.Case $2$: $(x + 2)^2 = -8(y - 1) \implies x^2 + 4x + 4 = -8y + 8 \implies x^2 + 4x + 8y - 4 = 0$.Both equations are valid,so the correct option is $(D)$.

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

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