Find the equation of parabola with vertex (-2,3), the equation directrix x+5=0
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The directrix is a vertical line x=-5, so this is a "sideways" parabola of the form:

x-h=a(y-k)2 where V(h,k) is the vertex, so h=-2 and y=3, and a is a constant.

x+2=a(y-3)2. The directrix cannot intersect the parabola so, since the vertex is at V(-2,3), and x=-5 is to the left of the vertex, a>0.

The axis of symmetry is horizontal line y=3.

All points on the parabola must be the same distance from the focus as they are from the directrix.

The point P(x,y) represents any point on the parabola. The focus lies on the axis of symmetry y=3, and V lies midway between F and the directrix line.

Since V is -2-(-5)=3 units horizontally from the directrix, the focus must be 3 units beyond V at F(-2+3,3)=F(1,3).

If P on the parabola is directly above the focus it's at the point P(1,y) which is 1-(-5)=6 units from the directrix line, so it must also be 6 units above F, that is, at P(1,3+6)=P(1,9).

Plug point P into the equation x+2=a(y-3)2 and we get 3=36a, making a=1/12.

Equation of the parabola is x+2=(y-3)2/12. The parabola is shown in red in the picture, with the points P and V labelled. The focus is at F and the directrix line is shown in blue. The axis of symmetry is shown in green. The equation can be written in various ways:

12x+24=y2-6y+9, 12x=y2-6y-15, y2-12x-6y-15=0.

Picture by courtesy of Desmos.

by Top Rated User (1.2m points)

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