body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

Method

Writing the Equation of a Parabola

The equation of a parabola can be determined if its focus and directrix is known.

Given any point on the parabola, the distance from that point to the focus is equal to the closest point on the directrix. Then, by using the distance formula, the lengths can be compared to get the equation for the parabola.

1

Draw line segments

Any point on a parabola has the same distance to the focus, $F,$ as to the closest point, $D,$ on the directrix. Thus, line segments are drawn from an arbitrary point, $P,$ on the parabola, to $F$ and $D .$

2

Use the distance formula

The length of each segment can be expressed with the distance formula since it's the distance between two points.

P D = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} P F = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

3

Substitute points

The coordinates for the focus can be read from the coordinate plane. Since the arbitrary point is any point on the parabola, its coordinates are unknown. Thus, the point

D,

on the directrix, shares the same

x

-value as

P

and has the

y

-value

1 .

F (- 2, - 5) P (x, y) D (x, 1)

The points can now be substituted into the expressions for the lengths.

P D = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(x, y)$ & $(x, 1)$

P D = (x - x)^{2} + (y - 1)^{2}

$x - x = 0$

P D = 0^{2} + (y - 1)^{2}

Calculate power

P D = (y - 1)^{2}

The same thing is now done for the length

P F .

P F = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(x, y)$ & $(- 2, - 5)$

P F = (x - (- 2))^{2} + (y - (- 5))^{2}

$- (- a) = a$

P F = (x + 2)^{2} + (y + 5)^{2}

The lengths are now expressed as:

P D P F = (y - 1)^{2} = (x + 2)^{2} + (y + 5)^{2}

4

Compare line segments

The point

P

are equidistant to

D

and

F,

that is,

P D

and

P F

has the same length. Thus, they can be put equal to each other.

(y - 1.5)^{2} = (x + 2)^{2} + (y + 4.5)^{2}

The equation for the parabola can now be found by solving the equation for

y .

(y - 1.5)^{2} = (x + 2)^{2} + (y + 4.5)^{2}

$LHS^{2} = RHS^{2}$

(y - 1.5)^{2} = (x + 2)^{2} + (y + 4.5)^{2}

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

y^{2} - 3 y + 1 . 5^{2} = (x + 2)^{2} + (y + 4.5)^{2}

ExpandPosPerfectSquare

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$

y^{2} - 3 y + 1 . 5^{2} = (x + 2)^{2} + y^{2} + 9 y - 4 . 5^{2}

Note that it is possible to expand

(x + 2)^{2}

but it's easier to leave it as the equation is solved for

y .

y^{2} - 3 y + 1 . 5^{2} = (x + 2)^{2} + y^{2} + 9 y - 4 . 5^{2}

▼

Solve for

y

$LHS - 1 . 5^{2} = RHS - 1 . 5^{2}$

y^{2} - 3 y = (x + 2)^{2} + y^{2} + 9 y - 4 . 5^{2} - 1 . 5^{2}

$LHS - y^{2} - 9 y = RHS - y^{2} - 9 y$

y^{2} - 3 y - y^{2} - 9 y = (x + 2)^{2} + 9 y + 4 . 5^{2}

CommutativePropAdd

Commutative Property of Addition

y^{2} - y^{2} - 3 y - 9 y = (x + 2)^{2} + 9 y + 4 . 5^{2}

Subtract terms

- 12 y = (x + 2)^{2} + 4 . 5^{2} - 1 . 5^{2}

Calculate power

- 12 y = (x + 2)^{2} + 20.25 - 2.25

Subtract term

- 12 y = (x + 2)^{2} + 18

$LHS / - 12 = RHS / - 12$

y = - \frac{( x + 2 ) ^{2}}{1 2}

Therefore, the equation for the parabola is

y = - \frac{( x + 2 ) ^{2} + 1 8}{1 2} .

Exercises

Recommended exercises

To get personalized recommended exercises, please login first.

Return to lesson.