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Conics: Parabolas:
Finding the Equation from Information
(page 3 of 4)

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.

• Write an equation for the parabola with focus at (0, –2) and directrix x = 2.
• 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,
–2)
. The absolute value of p is the distance between the vertex and the focus and the distance between the vertex and the directrix. (The sign on p tells me which way the parabola faces.) Since the focus and directrix are two units apart, then this distance has to be one unit, so | p | = 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)

• Write an equation of the parabola with vertex (3, 1) and focus (3, 5).
• Since the x-coordinates 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 right-side up parabola and p is positive. Since the vertex and focus are 5 – 1 = 4 units apart, then p = 4. Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved

And that's all I need for my equation, since they already gave me the vertex.

(xh)2 = 4p(yk)
(x – 3)2 = 4(4)(y – 1)

(x – 3)2 = 16(y – 1)

• Write an equation for the parabola with vertex (5, –2) and directrix y = –5.
• 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 right-side up parabola, so p is positive: p = 3. And that's all I need to find my equation:

(xh)2 = 4p(yk)
(x – 5)2 = 4(3)(y – (–2))

(x – 5)2 = 12(y + 2)

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 Cite this article as: Stapel, Elizabeth. "Conics: Parabolas: Finding the Equation From Information." Purplemath.     Available from https://www.purplemath.com/modules/parabola3.htm.     Accessed [Date] [Month] 2016

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