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=-20y#

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

The focus is #F=(0,-5)=(0,-p/2)#

So the directrix is #y=5#

Therefore,

#p=10#

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

#(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]}