Find the equation of the parabola with vertex | vedclass

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Question

(A) The axis of the parabola is the $x$-axis,as it passes through the vertex $(-3, 0)$ and is perpendicular to the directrix $x + 5 = 0$.Let the focus

Vedclass · Accepted Answer

$y^2 = 8(x + 3)$

Vedclass · Answer

(A) The axis of the parabola is the $x$-axis,as it passes through the vertex $(-3, 0)$ and is perpendicular to the directrix $x + 5 = 0$.Let the focus of the parabola be $S(a, 0)$. The vertex $(-3, 0)$ is the midpoint of the segment joining the focus $S(a, 0)$ and the point $Z(-5, 0)$ on the directrix.$-3 = \frac{a - 5}{2}$$-6 = a - 5$$a = -1$Thus,the focus is $S(-1, 0)$.Using the definition of a parabola,the distance from any point $P(x, y)$ on the parabola to the focus equals the distance to the directrix:$\sqrt{(x + 1)^2 + (y - 0)^2} = |x + 5|$$(x + 1)^2 + y^2 = (x + 5)^2$$x^2 + 2x + 1 + y^2 = x^2 + 10x + 25$$y^2 = 8x + 24$$y^2 = 8(x + 3)$

Find the equation of the parabola with vertex $(-3, 0)$ and directrix $x + 5 = 0$.

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