# 5.2: The Equation of the Parabola

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The equation of the parabola is often given in a number of different forms. One of the simplest of these forms is:

\[

(x-h)^{2}=4 p(y-k)

\]

A parabola is defined as the locus (or collection) of points equidistant from a given point (the focus) and a given line (the directrix). Another important point is the vertex or turning point of the parabola. If the equation of a parabola is given in standard form then the vertex will be \((h, k) .\) The focus will be a distance of \(p\) units from the vertex within the curve of the parabola and the directrix will be a distance of \(p\) units from the vertex outside the curve of the parabola. This value \((p)\) is called the focal distance.

Any point on the curve of the parabola is equidistant from the focus \((h, k+p)\) and the directrix \((h, k-p) .\) Notice that the focus is a point and is identified with the coordinates of the point while the directrix is a line and is identified with the equation for that line.

We can derive the standard equation for a parabola using the distance formula.

In the picture above the two distances labeled " \(d^{\prime \prime}\) should be the same distance. The vertical distance between the point \((x, y)\) and the directrix \(y=k-p\) is simply the difference between their \(y\) coordinates:

\[

d=y-(k-p)

\]

To find the distance between the point \((x, y)\) and the focus \((h, k+p)\) we need to use the distance formula:

\[

d=\sqrt{(x-h)^{2}+(y-(k+p))^{2}}

\]

Then we set the two distances equal to each other:

\[

\sqrt{(x-h)^{2}+(y-(k+p))^{2}}=y-(k-p)

\]

Square both sides:

\[

(x-h)^{2}+(y-(k+p))^{2}=(y-(k-p))^{2}

\]

We'll need to expand each side and collect like terms, but we'll leave the \((x-h)^{2}\) alone because it will appear in this form in the final equation.

\[

(x-h)^{2}+y^{2}-2(k+p) y+(k+p)^{2}=y^{2}-2(k-p) y+(k-p)^{2}

\]

The \(y^{2}\) terms cancel each other out:

\[

(x-h)^{2}-2(k+p) y+(k+p)^{2}=-2(k-p) y+(k-p)^{2}

\]

Now we'll expand the \(k, p\) and \(y\) terms:

\[

(x-h)^{2}-2 k y-2 p y+k^{2}+2 p k+p^{2}=-2 k y+2 p y+k^{2}-2 p k+p^{2}

\]

Here, the \(k^{2}\) and \(p^{2}\) terms cancel out, as do the \(-2 k y\) terms leaving:

\[

(x-h)^{2}-2 p y+2 p k=2 p y-2 p k

\]

If we collect everything excep the \((x-h)^{2}\) on the right hand side, we'll have:

\[

(x-h)^{2}=4 p y-4 p k

\]

Factor out \(4 p\) and we have the standard equation for a parabola:

\[

(x-h)^{2}=4 p(y-k)

\]

This equation will be different depending on the orientation of the parabola. An upward facing parabola will have this standard equation and both sides will have the same sign. For example, \((x-5)^{2}=12(y-1)\) is an upward facing parabola, as is \(-(x-5)^{2}=-12(y-1) .\) You can see that this is the same equation, but has been multiplied by -1 on both sides.

A parabola in which one side is positive and one side negative, like

\[

(x+1)^{2}=-8(y-10)

\]

is a downward facing parabola. This form might also appear as

\[

-(x+1)^{2}=8(y-10)

\]

The picture below illustrates this situation:

Notice that, in this case, the focus is below the vertex and the directrix is above it. Remember that the focus is always in the interior of the curve, while the directrix is always outside of the curve.

Parabolas can also open to the right or left. In these cases, it is the \(y\) variable that is squared in the standard equation:

\[

(y-k)^{2}=4 p(x-h)

\]

In these situations, the focus is to the left or right of the vertex. If both sides of the equation have the same \(\operatorname{sign},\) such as \((y-5)^{2}=12(x-1)\) or \(-(y-5)^{2}=-12(x-1)\) then this would be a parabola opening to the right.

If the two sides have opposite signs, then the parabola will open to the left

The orientation of the vertex, focus and directrix allows us to determine the equation of a parabola if we are given certain pieces of information about the vertex, focus and directrix. The vertex always falls halfway between the focus and directrix. The key pieces of information in determining the equation of a parabola are:

1) the vertex: this gives us the values for \(h\) and \(k\) for the equation.

2) the orientation: this allows us to determine the appropriate form of the equation

a) \(\quad(x-h)^{2}=4 p(y-k)\)

b) \(\quad(x-h)^{2}=-4 p(y-k)\)

c) \(\quad(y-k)^{2}=4 p(x-h)\)

d) \(\quad(y-k)^{2}=-4 p(x-h)\)

Given the following information determine an equation for the parabola described.

Focus at (3,2)\(\quad\) Vertex at (1,2)

First draw a little sketch of the problem:

since the focus always falls within the interior of the parabola's curve, this parabola is facing to the right. Its directrix is the line \(x=-1\)

The equation for this parabola would be:

\[

(y-2)^{2}=8(x-1)

\]

Exercises \(5.2(a)\)

Determine the equation in standard form for each of the parabolas described below.

1) Focus: (2,5) & Vertex: (2,6)

2) Focus: (-4,3) & Vertex: (-3,3)

3) Focus: (1.5,0) & Vetrex: (0,0)

4) Focus: (-9,0) & Vertex: (-9,-0.5)

5) Focus: (5,1) & Directrix: x=12

6) Focus: (0,3) & Directrix: y=1

7) Focus: (1,-3) & Directrix: y=2

8) Focus: (-4,8) & Directrix: x=-6

9) Vertex: (1,3), axis of symmetry parallel to \(y\) -axis

contains the point (4,7.5)

10) Vertex: (4,2), axis of symmetry parallel to \(y\) -axis

contains the point (5,0)

11) Focus: (0,0) & Vertex: (0,-1.25)

12) Focus: (0,0) & Vertex: (0,2.5)

13) Focus: (2,7) & Directrix: x=0.5

14) Focus: (-3,0) & Directrix: x=-2

15) Vertex: (3,-2), axis of symmetry parallel to \(x\) -axis,

contains the point (7,3)

16) Vertex: (-6,-6), axis of symmetry parallel to \(y\) -axis,

contains the point (0,0)

If we are given the equation of a parabola and need to find the vertex, focus and directrix, it is often helpful to put the equation in standard form. This usually requires completing the square.

If we're given the equation:

\[

x^{2}+6 x-4 y+1=0

\]

then we know that this will be either an upward or downward facing parabola. Let's move everything away from the \(x\) terms but leave a space to complete the

square:

\[

x^{2}+6 x=4 y-1

\]

To complete the square on this, we need add 9 to both sides so that we can rewrite the left side as \((x+3)^{2}\)

\[

\begin{aligned}

x^{2}+6 x+9 &=4 y-1+9 \\

(x+3)^{2} &=4 y+8

\end{aligned}

\]

Then we can factor out the coefficient of the \(y\) variable (even if it doesn't factor evenly).

\[

(x+3)^{2}=4(y+2)

\]

So, this is an upward facing parabola with the vertex at the point (-3,-2) . To find the focus and directrix, we need to know the vlaue of \(p .\) since \(4 p=4,\) then we know that \(p=1 .\) This means that the focus will be 1 unit above the vertex at the point (-3,-1) and the directrix will be one unit below the vertex at the line y=-3.

\begin{array}{c}

\text { Vertex: } \quad(-3,-2) \\

\text { Focus: } \quad(-3,-1) \\

\text {Directrix: } \quad (y=-3)

\end{array}

Here's another example:

Express the equation in standard form and determine the vertex, focus and directrix of the following parabola.

\[

y^{2}-2 y-1+8 x=0

\]

We know that this parabola will be either right or left facing.

First, let's move all the terms not containing \(y\) to the right-hand side but leave space to complete the square.

\[

y^{2}-2 y=-8 x+1

\]

Add 1 to both sides to complete the square:

\[

\begin{aligned}

y^{2}-2 y+1 &=-8 x+1+1 \\

(y-1)^{2} &=-8 x+2

\end{aligned}

\]

Factor out the coefficient of the \(x\) variable:

\[

(y-1)^{2}=-8(x-0.25)

\]

So, this is a left-facing parabola with a vertex at the point \((0.25,1) .\) To find the focal distance, we say that \(4 p=-8,\) so \(p=-2\)

since it is left-facing, the focus will be a distance of 2 units to the left of the vertex at the point (-1.75,1) and the directrix will be a distance of 2 units to the right of the vertex at the line \(x=2.25\)

Exercises \(5.2(b)\)

Express each equation in standard form and determine the vertex, focus and directrix of each parabola.

Standard form:

\[

\begin{array}{l}

4 p(y-k)=(x-h)^{2} \\

4 p(x-h)=(y-k)^{2}

\end{array}

\]

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

2) \(\quad (x+5)^{2}=12(y-3)\)

3) \(\quad (y+1)^{2}=6(x-2)\)

4) \(\quad (y+4)^{2}=10(x+1)\)

5) \(\quad (x+3)^{2}=-5(y+2)\)

6) \(\quad (x-4)^{2}=-7(y-6)\)

7) \(\quad (y+8)^{2}=-6(x+4)\)

8) \(\quad (y-3)^{2}=-9(x+4)\)

9) \(\quad (x^{2}+8 x+y+6=0\)

10) \(\quad x^{2}+6 x+y-3=0\)

11) \(\quad y^{2}+6 y+8 x+1=0\)

12) \(\quad y^{2}+8 y-4 x+8=0\)

13) \(\quad x^{2}+4 x-3 y+7=0\)

14) \(\quad x^{2}+2 x-6 y-11=0\)

15) \(\quad y^{2}+6 y-4 x+4=0\)

16) \(\quad y^{2}-4 y+3 x+9=0\)

17) \(\quad x+y^{2}-3 y+1=0\)

18) \(\quad 10+x+y^{2}+5 y=0\)

19) \(\quad x+y^{2}-3 y+4=0\)

20) \(\quad 3 x+y^{2}+8 y+4=0\)

21) \(\quad x^{2}+3 x+3 y-1=0\)

22) \(\quad x^{2}+5 x-4 y-1=0\)

23) \(\quad x^{2}-8 x-4 y+3=0\)

24) \(\quad 6 x-y^{2}-12 y+4=0\)

25) \(\quad 2 x+4 y^{2}+8 y-5=0\)

26) \(\quad 4 x^{2}-12 x+12 y+7=0\)

27) \(\quad 3 x^{2}-6 x-9 y+4=0\)

28) \(\quad 2 x-3 y^{2}+9 y+5=0\)

Write an equation in standard form for each parabola pictured below.