** CONIC SECTIONS
- THE PARABOLA**

Example 1:

Find the coordinates of the focus and the equation of the directrix of the parabola given by

.

First, we will convert the equation of the parabola to standard form.

Since there is a

y-term raised to the second power we know that this parabola is of the form .NOTE: The vertex

of this parabola is at(h, k)because there is only a(0, 0)y-term and no^{2}y-term.Let's multiply both sides of the original equation by

to change the coefficient of8toy^{2}.1Now, we can write the equation as follows:

The standard form contains a

. Currently,4psits in its position. Therefore, we can say that8

8= 4pand

p =2We can now write the standard form of the original parabolic equation as follows:

where

,h = 0, andk = 0p =2

Coordinates of the Focus(h + p, k)

(02, 0) =(2, 0)

Equation of the Directrixx= hp

x= 0(2)

x= 2

Example 2:

Find the coordinates of the focus and the equation of the directrix of the parabola given by

.

First, we will convert the equation of the parabola to standard form.

Since there is an

x-term raised to the second power we know that this parabola is of the form .NOTE: The vertex

of this parabola is at(h, k)because there is only an(0, 0)x-term and no^{2}x-term.Let's divide both sides of the original equation by

to to change the coefficient of4tox^{2}.1Now, we can write the equation as follows:

The standard form contains a

. Currently, sits in its position. Therefore, we can say that4pand

We can now write the standard form of the original parabolic equation as follows:

where

,h = 0, andk = 0

Coordinates of the Focus(h, k + p)

^{=}

Equation of the Directrixy = kp

and

Example 3:

Find the coordinates of the vertex and the focus and the equation of the directrix of the parabola given by

.

First, we will convert the equation of the parabola to standard form.

Since there is an

x-term raised to the second power we know that this parabola is of the form .Let's first multiply both sides of the original equation by

to change the coefficient of2tox^{2}.1Then .

Next, we'll isolate the terms containing

on one side of the equation as follows.x

We need

(xon the right! We will use theh)^{2}Square Completion Methodto alter the right side appropriately.We will utilize the term

. We divide its coefficient2xby2, then we raise this quotient to the second power,2That is, .

Now, we add this

to1BOTHsides of the equation .

1 2y + 1 = x^{2}+ 2x + 1and

Continuing, we get

22y=.(x + 1)^{2}Now we can begin to rewrite this equation into standard form .

We have

, but we can write this as(x + 1)^{2}[x (1)].This is^{2}(x.h)^{2}We have

2, but we can write this as2y2(y.1)The standard form contains a

. Currently,4psits in its position. Therefore, we can say that2

2 = 4pthen

p =We can now write the standard form of the original parabolic equation as follows:

[x (1)]^{2}= 4()(ywhere1)h =,2, andk = 1p =

Coordinates of the Vertex(h, k)

(1, 1)

Coordinates of the Focus(h, k + p)

(1, 1 )=(1, )

Equation of the Directrixy = k p

y = 1( ) = 1 +and

Example 4:

Find the coordinates of the vertex and focus and the equation of the directrix of the parabola given by

.

First, we will convert the equation of the parabola to standard form.

Since there is a

y-term raised to the second power we know that this parabola is of the form .Let's first multiply both sides of the original equation by

to change the coefficient of4toy^{2}.1Then .

Next, we'll isolate the terms containing

on one side of the equation as follows.yWe need

on the right! We will use the(yk)^{2}Square Completion Methodto alter the right side appropriately.We will utilize the term

. We divide its coefficient4yby4, then we raise this quotient to the second power,2That is, .

Now, we add this

to4BOTHsides of the equation .

4x 16 + 4 = y^{2}4y + 4and

4x 12 =y^{2}4y + 4Continuing, we get

4x 12 =.(y2)^{2}Now we can begin to rewrite this equation into standard form .

We have

which is(y2)^{2}.(yk)^{2}We have

, but we can write this as4x 12.4(x 3)The standard form contains a

. Currently,4psits in its position. Therefore, we can say that4

4 = 4pthen

p = 1We can now write the standard form of the original parabolic equation as follows:

where

,h = 3, andk = 2p = 1

Coordinates of the Focus(h + p, k)

(3 + 1, 2) =(4, 2)

Equation of the Directrixx= hp

x= 31and

x= 2