** CONIC SECTIONS
- THE ELLIPSE**

Example 1:

Find the coordinates of the center, the vertices, and foci of the ellipse given by .

This equation is almost in standard form. All we have to do is the following:

where

,h =4,k =3(from larger denominator), anda = 3b = 2Now, we can find the requested information as follows:

Coordinates of the Center(h, k)

(4,3)

Coordinates of the Vertices(h + a, k)and(h a, k)

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

(43, 3) = (7, 3)

Coordinates of the Foci(h + c, k)and,(hc, k)whereSince

anda = 3, then .b = 2and

Example 2:

Find the coordinates of the center, vertices, and foci of the ellipse given by .

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

It is either or , where

First, we will divide both sides of the equation by

to get16

We can now write the standard form of the given elliptic equation as follows:

where

,h =1,k =2(from larger denominator), anda = 4b = 2Now, we can find the requested information as follows:

Coordinates of the Center(h, k)

(1,2)

Coordinates of the Vertices(h, k + a)and(h, ka)

and(1, 2 + 4)(1, 24)

and(1, 2)(1, 6)

Coordinates of the Foci(h, k + c)and,(h, kc)whereSince

anda = 4, then .b = 2and

Example 3:

Find the coordinates of the center, vertices, and foci of the ellipse given by

.

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

NOTE: The center

of the given ellipse is at(h, k)because there is only an(0, 0)x-term and a^{2}y-term. Therefore, we can write^{2}where

,h = 0,k = 0(from larger denominator), anda = 5b = 4Now, we can find the requested information as follows:

Coordinates of the Center(h, k)

(0, 0)

Coordinates of the Vertices(h, k + a)and(h, ka)

and(0, 0 + 5)(0, 05)

and(0, 5)(0, 5)

Coordinates of the Foci(h, k + c)and,(h, kc)whereSince

anda = 5, then .b = 4

and(0, 0 + 3)(0, 03)

and(0, 3)(0, 3)

Example 4:

Find the coordinates of the center, vertices, and foci of the ellipse given by .

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

It is either or , where

First, we will divide both sides of the equation by

to get36We can now write the standard form of the given elliptic equation as follows:

where

,h =0,k =0(from larger denominator), anda = 3b = 2Now, we can find the requested information as follows:

Coordinates of the Center(h, k)

(0, 0)

Coordinates of the Vertices(h, k + a)and(h, ka)

and(0, 0 + 3)(0, 03)

and(0, 3)(0, 3)

Coordinates of the Foci(h, k + c)and,(h, kc)whereSince

anda = 3, then .b = 2and

and

Example 5:

Match the names hyperbola, ellipse, parabola, and circle to the following conic equations.

a.

We must recognize this as a hyperbola because of its general characteristics. There is a squared

xandy-term and one of them is negative. It is actually ahyperbolawith center at the origin.b.

We must recognize this as an

ellipsebecause of its general characteristics. There is a squaredxandy-term and both are positive. Also, the squared terms havedifferentdenominators.c.

We must recognize this as an

ellipsebecause of its general characteristics. There is a squaredxandy-term and both are positive. Also, the squared terms havedifferentcoefficients (or denominators).d.

We must recognize this as a

circlebecause of its general characteristics. There is a squaredxandy-term and both are positive. Also, the squared terms have thesamedenominators.e.

We must recognize this as a

circlebecause of its general characteristics. There is a squaredxandy-term and both are positive. Also, the squared terms have thesamecoefficients (or denominators), namely.1f.

We must recognize this as the standard form of a

parabolabecause of its general characteristics. There is only one squared term, namelyy^{2}.g.

We must recognize this as a

hyperbolabecause of its general characteristics. There is a squaredxandy-term and one of them is negative.