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Your homework assignment is to draw the ellipse
16(x−2)2+4(y+3)2=144
. What is the vertex of your graph and where will the foci of the ellipse be located?
Ellipses Centered at (h,k)
An ellipse does not always have to be placed with its center at the origin. If the center is
(h,k)
the entire ellipse will be shifted
h
units to the left or right and
k
units up or down. The equation becomes
(x−h)2a2+(y−k)2b2=1
. We will address how the vertices, co-vertices, and foci change in the following problem.
Let's graph
(x−3)216+(y+1)24=1
. Then, we'll find the vertices, co-vertices, and foci.
First, we know this is a horizontal ellipse because
16>4
. Therefore, the center is
(3,−1)
and
a=4
and
b=2
. Use this information to graph the ellipse.
To graph, plot the center and then go out 4 units to the right and left and then up and down two units. This is also how you can find the vertices and co-vertices. The vertices are
(3±4,−1)
or
(7,−1)
and
(−1,−1)
. The co-vertices are
(3,−1±2)
or
(3,1)
and
(3,−3)
.
[Figure 1]
To find the foci, we need to find
c
using
c2=a2−b2
.
c2c=16−4=12=23–√
Therefore, the foci are
(3±23–√,−1)
.
From this problem, we can create formulas for finding the vertices, co-vertices, and foci of an ellipse with center
(h,k)
. Also, when graphing an ellipse, not centered at the origin, make sure to plot the center.
Orientation Equation Vertices Co-Vertices Foci
Horizontal
(x−h)2a2+(y−k)2b2=1
(h±a,k)
(h,k±b)
(h±c,k)
Vertical
(x−h)2b2+(y−k)2a2=1
(h,k±a)
(h±b,k)
(h,k±c)
Now, let's find the equation of the ellipse with vertices
(−3,2)
and
(7,2)
and co-vertex
(2,−1)
.
These two vertices create a horizontal major axis, making the ellipse horizontal. If you are unsure, plot the given information on a set of axes. To find the center, use the midpoint formula with the vertices.
(−3+72,2+22)=(42,42)=(2,2)
The distance from one of the vertices to the center is
a
,
|7−2|=5
. The distance from the co-vertex to the center is
b
,
|−1−2|=3
. Therefore, the equation is
(x−2)252+(y−2)232=1
or
(x−2)225+(y−2)29=1
.
Finally, let's graph
49(x−5)2+25(y+2)2=1225
and find the foci.
First we have to get this into standard form, like the equations above. To make the right side 1, we need to divide everything by 1225.