How do you find the standard form of the equation of the parabola with the given characteristic and vertex at the origin and Focus: (0, −5)?

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Answer 1 · 2017-01-07T07:12:20

The equation is $x^{2} = - 20 y$ $x^2=-20y$

The vertex is $V = (0, 0)$ $V=(0,0)$

The focus is $F = (0, - 5) = (0, - \frac{p}{2})$ $F=(0,-5)=(0,-p/2)$

So the directrix is $y = 5$ $y=5$

Therefore,

$p = 10$ $p=10$

Any point on the parabola is equidistant from the focus and the directrix

$(5 - y) = \sqrt{x^{2} + {(y + 5)}^{2}}$ $(5-y)=sqrt(x^2+(y+5)^2)$

$cancel25+cancely^2-10y=x^2+cancely^2+10y+cancel25$

The equation of the parabola is

$x^2=-20y$

graph{(x^2+20y)(y-5)((x)^2+(y+5)^2-0.01)=0 [-12.66, 12.65, -6.33, 6.33]}