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Start by identifying whether you have a horizontal or a vertical ellipse.
Equation: (x-2)^29+ (y-5)^24=1
Graph:
Let's start by recalling the standard form for horizontal and vertical ellipses and highlight some of their important characteristics.
| Horizontal Ellipse | Vertical Ellipse | |
|---|---|---|
| Standard Form | (x- h)^2/a^2+(y- k)^2/b^2=1, a and b positive, with a> b |
(x- h)^2/b^2+(y- k)^2/a^2=1, a and b positive, with a> b |
| Center | ( h, k) | ( h, k) |
| Vertices | ( h± a, k) | ( h, k± a) |
| Co-Vertices | ( h, k± b) | ( h± b, k) |
| Length of Major Axis | 2 a units | 2 a units |
| Length of Minor Axis | 2 b units | 2 b units |
| Foci | ( h± c, k), c^2= a^2- b^2 |
( h, k± c), c^2= a^2- b^2 |
We are given the center, vertices, and co-vertices of an ellipse, and we want to write its equation. Note that the center has the same x-coordinate as the co-vertices and the same y-coordinate as the vertices. According to our table, this corresponds to a horizontal ellipse.
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Center & Vertices & Co-vertices
( 2, 5)& (5, 5)and(- 1, 5)& ( 2,3)and ( 2,7)
Following the same procedure, we will use one of the co-vertices and the center to find the positive value of b.
Knowing that a= 3 and b= 2, we now have everything we need to write the equation of the ellipse. (x- 2)^2/3^2+(y- 5)^2/2^2=1 ⇕ (x-2)^2/9+(y-5)^2/4=1 Now, let's draw our ellipse. To do so, we will first isolate the y-variable.
Now, we will make a table of values to find points on the curve. Be aware that the radicand cannot be negative!
| x | ± 2sqrt(1 - (x-2)^2/9)+5 | y=2sqrt(1 - (x-2)^2/9)+5 | y=- 2sqrt(1 - (x-2)^2/9)+5 |
|---|---|---|---|
| - 1 | ± 2sqrt(1 - ( - 1-2)^2/9)+5 | 5 | 5 |
| 0 | ± 2sqrt(1 - ( 0-2)^2/9)+5 | ≈ 6.5 | ≈ 3.5 |
| 1 | ± 2sqrt(1 - ( 1-2)^2/9)+5 | ≈ 6.9 | ≈ 3.1 |
| 2 | ± 2sqrt(1 - ( 2-2)^2/9)+5 | 7 | 3 |
| 3 | ± 2sqrt(1 - ( 3-2)^2/9)+5 | ≈ 6.9 | ≈ 3.1 |
| 4 | ± 2sqrt(1 - ( 4-2)^2/9)+5 | ≈ 6.5 | ≈ 3.5 |
| 5 | ± 2sqrt(1 - ( 5-2)^2/9)+5 | 5 | 5 |
Finally, let's plot the points and connect them with a smooth curve.