# How do you find the equation of a parabola with vertex at the origin and directrix x=-2?

##y^2=8x##

The measure shape of the parabola is ##y^2=4ax##, giving a parabola after a while its axis congruous to the ##x##-axis, vertex at the rise, standsummit ##(a,0)## and directrix ##x=-a##. So in your subject ##a=2##, giving ##y^2=4ax##.

Alternatively, you can from a specification of a parabola, which is the set of all summits ##(x,y)## such that the interval from the summit to the directrix ##x=-2## is the selfselfsame as the interval to the standsummit (2,0). (The vertex is half-way among the standsummit and the directrix.)

##(x-(-2))^2=(x-a)^2+y^2## ##cancel(x^2)+4ax+expunge 4=cancel(x^2)-4ax+expunge 4+y^2## ##y^2=4ax+4ax=8ax##

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