The equation of the parabola whose vertex is | vedclass

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(A) Since the axis of the parabola is vertical,the equation is of the form $(x - h)^2 = 4a(y - k)$,where $(h, k)$ is the vertex.Given vertex $(h, k) =

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${x^2} + 2x - 2y - 3 = 0$

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(A) Since the axis of the parabola is vertical,the equation is of the form $(x - h)^2 = 4a(y - k)$,where $(h, k)$ is the vertex.Given vertex $(h, k) = (-1, -2)$,the equation becomes $(x + 1)^2 = 4a(y + 2)$.Since the parabola passes through the point $(3, 6)$,we substitute these coordinates into the equation:$(3 + 1)^2 = 4a(6 + 2)$$4^2 = 4a(8)$$16 = 32a$$a = \frac{16}{32} = \frac{1}{2}$.Substituting $a = \frac{1}{2}$ back into the equation:$(x + 1)^2 = 4(\frac{1}{2})(y + 2)$$(x + 1)^2 = 2(y + 2)$$x^2 + 2x + 1 = 2y + 4$$x^2 + 2x - 2y - 3 = 0$.

The equation of the parabola whose vertex is $(-1, -2)$,axis is vertical and which passes through the point $(3, 6)$,is

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