What is the standard form of the equation of the parabola with a directrix at x=4 and a focus at (-7,-5)?

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Answer 1 · 2017-12-25T12:22:18

The standard equation of parabola is $(y+5.5)^2 = -22(x+1.5)$

Focus is at $(-7,-5)$ and directrix is $x=4$ . Vertex is at midway

between focus and directrix. Therefore vertex is at

$((-7+4)/2,-5) or (-1.5,-5)$ The equation of horizontal

parabola opening left is

$(y-k)^2 = -4p(x-h) ; h=-1.5 ,k=-5$

or $(y+5.5)^2 = -4p(x+1.5)$ . The distance between focus and

vertex is $p=7-1.5=5.5$ . Thus the standard equation of

horizontal parabola is $(y+5.5)^2 = -4*5.5(x+1.5)$ or

$(y+5.5)^2 = -22(x+1.5)$

graph{(y+5.5)^2=-22(x+1.5) [-160, 160, -80, 80]}