The equation of the parabola with (-3, 0) as | vedclass

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(A) The focus of the parabola is $S = (-3, 0)$ and the directrix is $x + 5 = 0$.Since the directrix is a vertical line $(x = -5)$,the axis of the para

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$y^2 = 4(x + 4)$

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(A) The focus of the parabola is $S = (-3, 0)$ and the directrix is $x + 5 = 0$.Since the directrix is a vertical line $(x = -5)$,the axis of the parabola is the $x$-axis $(y = 0)$.The vertex is the midpoint of the focus and the point where the axis intersects the directrix.The intersection of the axis $(y = 0)$ and the directrix $(x = -5)$ is $(-5, 0)$.Vertex $V = \left( \frac{-3 + (-5)}{2}, 0 \right) = (-4, 0)$.The distance from the vertex to the focus is $a = -3 - (-4) = 1$.The standard form of a parabola opening to the right is $(y - k)^2 = 4a(x - h)$,where $(h, k)$ is the vertex.Substituting $h = -4, k = 0$,and $a = 1$,we get $(y - 0)^2 = 4(1)(x - (-4))$.Thus,the equation is $y^2 = 4(x + 4)$.

The equation of the parabola with $(-3, 0)$ as focus and $x + 5 = 0$ as directrix is

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