Conics:
Parabolas: Sections: Introduction, Finding information from the equation, Finding the equation from information, Word problems & Calculators You will also need to work the other way, going from the properties of the parabola to its equation.
The vertex is always halfway between the focus and the directrix, and the parabola always curves away from the directrix, so I'll do a quick graph showing the focus, the directrix, and a rough idea of where the parabola will go:
So the vertex, exactly between the focus
and directrix, must be at (h,
k) = (1, Since the focus is to the left of the
vertex and directrix, then the parabola faces left (as I'd shown in
my picture) and I get a negative value for p:
p = –1.
Since this is a "sideway" parabola, then the y
part gets squared, rather than the x
part. So the conics form of the equation must be: (y – (–2))^{2} = 4(–1)(x – 1), or (y + 2)^{2} = –4(x – 1)
Since the xcoordinates of the vertex and focus are the same, they are one of top of the other, so this is a regular vertical parabola, where the x part is squared. Since the vertex is below the focus, this is a rightside up parabola and p is positive. Since the vertex and focus are 5 – 1 = 4 units apart, then p = 4. Copyright © Elizabeth Stapel 20102011 All Rights Reserved And that's all I need for my equation, since they already gave me the vertex. (x – h)^{2}
= 4p(y – k)
The directrix is an horizontal line; since this line is perpendicular to the axis of symmetry, then this must be a regular parabola, where the x part is squared. The distance between the vertex and the directrix is –5 – (–2) = –5 + 2 = 3. Since the directrix is below the vertex, then this is a rightside up parabola, so p is positive: p = 3. And that's all I need to find my equation: (x – h)^{2}
= 4p(y – k)
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