Lesson Objectives

- Learn how to find the foci of an ellipse
- Learn how to find the vertices and covertices of an ellipse
- Learn how to write the equation of an ellipse
- Learn how to sketch the graph of an ellipse

## How to Sketch the Graph of an Ellipse

In this lesson, we will continue to learn about conic sections with a focus on the ellipse. In the last lesson, we learned that the parabola was defined geometrically as the set of all points in a plane that are equidistant from a fixed point known as the focus and a fixed line known as the directrix. Similarly, an ellipse is defined geometrically as the set of all points in a plane the sum of whose distances from two fixed points is constant. The two fixed points are known as the foci, which is the plural form of focus.

Example #1: Find the vertices, covertices, and foci, and then sketch the graph. $$25x^2 + 4y^2 - 100 = 0$$ Add 100 to each side: $$25x^2 + 4y^2 = 100$$ Divide both sides by 100: $$\frac{25}{100}x^2 + \frac{4}{100}y^2 = \frac{100}{100}$$ Simplify: $$\frac{1}{4}x^2 + \frac{1}{25}y^2 = 1$$ $$\frac{x^2}{4} + \frac{y^2}{25} = 1$$ Notice that 25 is larger than 4, which means the major axis is vertical. $$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$$ $$a^2 = 25, b^2 = 4$$ Note: a, b, and c are positive as defined by our derivation above. As we previously mentioned in the tutorial, think of a as the distance from the center to each vertex, b as the distance from the center to each covertex, and c as the distance from the center to each focus. Take this into consideration when dealing with square roots in this section. $$a^2 = 25$$ $$a = \sqrt{25} = 5$$ $$b^2 = 4$$ $$b = \sqrt{4} = 2$$ $$c^2 = a^2 - b^2$$ $$c = \sqrt{a^2 - b^2}$$ $$c = \sqrt{25 - 4} = \sqrt{21}$$ We only want the principal square root in each case since a, b, and c each represent a distance.

Vertices: $$(0, -a), (0, a)$$ $$a = 5$$ $$(0, -5), (0, 5)$$ Covertices: $$(-b, 0), (b, 0)$$ $$b = 2$$ $$(-2, 0), (2, 0)$$ Foci: $$c = \sqrt{21}$$ $$(0, -c), (0, c)$$ $$(0, -\sqrt{21}), (0, \sqrt{21})$$ Desmos Link for More Detail Example #2: Write the equation of the ellipse in standard form and then sketch the graph.

Vertices: $$(10, 0), (-10, 0)$$ Foci: $$\left(4\sqrt{6}, 0\right), \left(-4\sqrt{6}, 0\right)$$ Recall that the vertices lie on the major axis. Since we have the same y-coordinate of 0 in each case and different x-coordinates (10 and -10), we can conclude that the major axis is horizontal. Additionally, we know the center is at the origin since the midpoint of the major axis is the center of the ellipse. $$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ Plug into the midpoint formula: $$M = \left(\frac{10 + (-10)}{2}, \frac{0 + 0}{2}\right) = (0, 0)$$ Center: $$(0, 0)$$ Here we will choose the formula for an ellipse with its center at the origin where the major axis is horizontal. $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ Recall that a can be thought of as the distance from the center to each vertex. Since the center is at (0, 0), and a vertex is given as (10, 0), we can conclude that a is 10. $$a = 10 - 0 = 10$$ Additionally, c can be thought of as the distance from the center to each focus. $$c = 4\sqrt{6} - 0 = 4\sqrt{6}$$ Since we need to get b We can use the following table when considering translations with ellipses.

If h and k are positive real numbers (h > 0 and k > 0):

Let's look at an example.

Example #3: Find the center, vertices, covertices, foci, and equation in standard form, and then sketch the graph. $$49x^2 + 9y^2 - 196x - 245 = 0$$ Group the x-terms together. $$49x^2 - 196x + 9y^2 - 245 = 0$$ Factor out the coefficient of x

Vertices: $$(2, -7), (2, 7)$$ To get the foci, recall that c is the distance from the center to each focus. $$c^2 = a^2 - b^2$$ $$a^2 = 49, b^2 = 9$$ Plug in: $$c^2 = 49 - 9 = 40$$ $$c = \sqrt{40} = 2\sqrt{10}$$ Foci: $$\left(2, 2\sqrt{10}\right), \left(2, -2\sqrt{10}\right)$$ To get the covertices, recall that b is the distance from the center to each covertex. $$b^2 = 9$$ $$b = 3$$ $$2 - 3 = -1, 2 + 3 = 5$$ Covertices: $$(-1, 0), (5, 0)$$ Desmos Link for More Detail

Let's look at an example.

Example #4: Find the standard form of the ellipse and sketch the graph.

Vertices: $$(-10, 8), (-10, -18)$$ Foci: $$(-10, 7), (-10, -17)$$ Recall that the vertices lie on the major axis. Since we have the same x-coordinate of -10 in each case and different y-coordinates (8 and -18), we can conclude that the major axis is vertical. The center is the midpoint of the major axis. $$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ Plug into the midpoint formula: $$M = \left(\frac{-10 + (-10)}{2}, \frac{8 + (-18)}{2}\right) = (-10, -5)$$ Center: $$(-10, -5)$$ Let's plug in what we know so far. $$\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1$$ $$\frac{(x + 10)^2}{b^2} + \frac{(y + 5)^2}{a^2} = 1$$ a is the distance from the center to each vertex. $$a = 8 - (-5) = 8 + 5 = 13$$ $$a^2 = 13^2 = 169$$ $$\frac{(x + 10)^2}{b^2} + \frac{(y + 5)^2}{169} = 1$$ We can get b

- An ellipse has two axes of symmetry
- Major axis (longer one)
- Minor axis (shorter one)

- The foci are located on the major axis
- The midpoint of the major axis is the center of the ellipse
- The axes are perpendicular at the center
- The endpoints of the major axis are the vertices of the ellipse
- The endpoints of the minor axis are the covertices of the ellipse
- The graph of an ellipse is not the graph of a function
- It does not pass the vertical line test

### Deriving the Standard Form of the Equation of an Ellipse

We can derive the equation of an ellipse using its geometric definition. The distance from V_{1}(a, 0) to F_{1}(c, 0) is (a - c) and the distance from V_{1}(a, 0) to F_{2}is (a + c). The sum of the distances is 2a. $$(a - c) + (a + c) = 2a$$ Recall from our definition of an ellipse that it is the set of all points in a plane the sum of whose distances from the foci is constant. Since the vertex V_{1}(a, 0) is on the ellipse, the sum of 2a can be used to set up the following. $$d(P, F_1) + d(P, F_2) = 2a$$ P(x, y) is any point on the ellipse. Let's plug into the distance formula. $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ $$d(P, F_1) = \sqrt{(x - c)^2 + (y - 0)^2}$$ $$=\sqrt{(x - c)^2 + y^2}$$ $$d(P, F_2) = \sqrt{(x - (-c))^2 + (y - 0)^2}$$ $$=\sqrt{(x + c)^2 + y^2}$$ $$d(P, F_1) + d(P, F_2) = 2a$$ Plug in: $$\sqrt{(x - c)^2 + y^2} + \sqrt{(x + c)^2 + y^2} = 2a$$ Isolate one of the radicals: $$\sqrt{(x + c)^2 + y^2} = 2a - \sqrt{(x - c)^2 + y^2}$$ Square both sides: $$\left(\sqrt{(x + c)^2 + y^2}\right)^2 = \left(2a - \sqrt{(x - c)^2 + y^2}\right)^2$$ $$(x + c)^2 + y^2 = (2a)^2 - (2)(2a)\left(\sqrt{(x - c)^2 + y^2}\right) + \left(\sqrt{(x - c)^2 + y^2}\right)^2$$ $$(x + c)^2 + y^2 = 4a^2 - 4a\sqrt{(x - c)^2 + y^2} + (x - c)^2 + y^2$$ Expand the squares: $$x^2 + 2cx + c^2 + y^2 = 4a^2 - 4a\sqrt{(x - c)^2 + y^2} + x^2 - 2cx + c^2 + y^2$$ $$2cx = 4a^2 - 4a\sqrt{(x - c)^2 + y^2} - 2cx$$ $$4cx = 4a^2 - 4a\sqrt{(x - c)^2 + y^2}$$ Divide both sides by 4: $$\require{cancel}\frac{\cancel{4}cx}{\cancel{4}} = \frac{\cancel{4}a^2}{\cancel{4}} - \frac{\cancel{4}a\sqrt{(x - c)^2 + y^2}}{\cancel{4}}$$ $$cx = a^2 - a\sqrt{(x - c)^2 + y^2}$$ Isolate the radical and square both sides: $$cx - a^2 = - a\sqrt{(x - c)^2 + y^2}$$ $$\left(cx - a^2\right)^2 = \left(-a\sqrt{(x - c)^2 + y^2}\right)^2$$ $$(cx)^2 - 2(cx)(a^2) + (a^2)^2 = (-a)^2 \cdot \left(\sqrt{(x - c)^2 + y^2}\right)^2$$ $$c^2x^2 - 2a^2cx + a^4 = a^2\left[(x - c)^2 + y^2\right]$$ $$c^2x^2 - 2a^2cx + a^4 = a^2\left(x^2 - 2cx + c^2 + y^2\right)$$ Distribute the a^{2}on the right: $$c^2x^2 - 2a^2cx + a^4 = a^2x^2 - 2a^2cx + a^2c^2 + a^2y^2$$ $$c^2x^2 + a^4 = a^2x^2 + a^2c^2 + a^2y^2$$ $$a^4 - a^2c^2 = a^2x^2 - c^2x^2 + a^2y^2$$ Switch sides: $$a^2x^2 - c^2x^2 + a^2y^2 = a^4 - a^2c^2$$ Factor: $$x^2(a^2 - c^2) + a^2y^2 = a^2(a^2 - c^2)$$ Divide both sides by a^{2}(a^{2}- c^{2}): $$\frac{x^2\cancel{(a^2 - c^2)}}{a^2\cancel{(a^2 - c^2)}} + \frac{\cancel{a^2}y^2}{\cancel{a^2}(a^2 - c^2)} = \frac{1\cancel{a^2}\cancel{(a^2 - c^2)}}{\cancel{a^2}\cancel{(a^2 - c^2)}}$$ $$\frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1$$ Since B_{1}(0, b) is on the ellipse. $$d(B_1, F_1) + d(B_1, F_2) = 2a$$ $$d(B_1, F_1) = \sqrt{(0 - c)^2 + (b - 0)^2}$$ $$=\sqrt{c^2 + b^2}$$ $$d(B_1, F_2) = \sqrt{(0 - (-c))^2 + (b - 0)^2}$$ $$=\sqrt{c^2 + b^2}$$ $$d(B_1, F_1) + d(B_1, F_2) = 2a$$ Plug in: $$\sqrt{c^2 + b^2} + \sqrt{c^2 + b^2} = 2a$$ $$2\sqrt{c^2 + b^2} = 2a$$ Divide both sides by 2: $$\frac{\cancel{2}\sqrt{c^2 + b^2}}{\cancel{2}} = \frac{\cancel{2}a}{\cancel{2}}$$ $$a = \sqrt{c^2 + b^2}$$ Square both sides: $$a^2 = c^2 + b^2$$ Subtract c^{2}from each side: $$a^2 - c^2 = b^2$$ Switch sides: $$b^2 = a^2 - c^2$$ Return to our formula: $$\frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1$$ Replace a^{2}- c^{2}with b^{2}: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ The above is the standard form of the equation of an ellipse with its center at the origin and its foci on the x-axis. We could use a similar process to derive the standard form of the equation of an ellipse with its center at the origin and its foci on the y-axis.### Standard Forms of the Equations of an Ellipse

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$- Center is at the origin
- Major axis on x-axis and has length 2a
- Vertices (±a, 0)
- Minor axis on y-axis and has length 2b
- Covertices (0, ±b)
- foci (±c, 0)

- Center is at the origin
- Major axis on y-axis and has length 2a
- Vertices (0, ±a)
- Minor axis on x-axis and has length 2b
- Covertices (±b, 0)
- foci (0, ±c)

^{2}will be associated with x^{2}. For the second scenario, the major axis is now on the y-axis (vertical), and the positive y-intercept is larger than the positive x-intercept. This means the a^{2}will be associated with y^{2}. An easy way to remember this is that the vertices are on the major axis and are a units away from the center, while the covertices are on the minor axis and are b units away from the center. Additionally, the foci are on the major axis, c units away from the center. Let's look at some examples.Example #1: Find the vertices, covertices, and foci, and then sketch the graph. $$25x^2 + 4y^2 - 100 = 0$$ Add 100 to each side: $$25x^2 + 4y^2 = 100$$ Divide both sides by 100: $$\frac{25}{100}x^2 + \frac{4}{100}y^2 = \frac{100}{100}$$ Simplify: $$\frac{1}{4}x^2 + \frac{1}{25}y^2 = 1$$ $$\frac{x^2}{4} + \frac{y^2}{25} = 1$$ Notice that 25 is larger than 4, which means the major axis is vertical. $$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$$ $$a^2 = 25, b^2 = 4$$ Note: a, b, and c are positive as defined by our derivation above. As we previously mentioned in the tutorial, think of a as the distance from the center to each vertex, b as the distance from the center to each covertex, and c as the distance from the center to each focus. Take this into consideration when dealing with square roots in this section. $$a^2 = 25$$ $$a = \sqrt{25} = 5$$ $$b^2 = 4$$ $$b = \sqrt{4} = 2$$ $$c^2 = a^2 - b^2$$ $$c = \sqrt{a^2 - b^2}$$ $$c = \sqrt{25 - 4} = \sqrt{21}$$ We only want the principal square root in each case since a, b, and c each represent a distance.

Vertices: $$(0, -a), (0, a)$$ $$a = 5$$ $$(0, -5), (0, 5)$$ Covertices: $$(-b, 0), (b, 0)$$ $$b = 2$$ $$(-2, 0), (2, 0)$$ Foci: $$c = \sqrt{21}$$ $$(0, -c), (0, c)$$ $$(0, -\sqrt{21}), (0, \sqrt{21})$$ Desmos Link for More Detail

$$\frac{x^2}{4} + \frac{y^2}{25} = 1$$

Vertices: $$(10, 0), (-10, 0)$$ Foci: $$\left(4\sqrt{6}, 0\right), \left(-4\sqrt{6}, 0\right)$$ Recall that the vertices lie on the major axis. Since we have the same y-coordinate of 0 in each case and different x-coordinates (10 and -10), we can conclude that the major axis is horizontal. Additionally, we know the center is at the origin since the midpoint of the major axis is the center of the ellipse. $$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ Plug into the midpoint formula: $$M = \left(\frac{10 + (-10)}{2}, \frac{0 + 0}{2}\right) = (0, 0)$$ Center: $$(0, 0)$$ Here we will choose the formula for an ellipse with its center at the origin where the major axis is horizontal. $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ Recall that a can be thought of as the distance from the center to each vertex. Since the center is at (0, 0), and a vertex is given as (10, 0), we can conclude that a is 10. $$a = 10 - 0 = 10$$ Additionally, c can be thought of as the distance from the center to each focus. $$c = 4\sqrt{6} - 0 = 4\sqrt{6}$$ Since we need to get b

^{2}, let's plug into our formula. $$b^2 = a^2 - c^2$$ $$a = 10, c = 4\sqrt{6}$$ $$b^2 = 10^2 - (4\sqrt{6})^2$$ $$b^2 = 100 - (16)(6)$$ $$b^2 = 100 - 96$$ $$b^2 = 4$$ $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ Plug in for a^{2}and b^{2}. $$a^2 = 100, b^2 = 4$$ $$\frac{x^2}{100} + \frac{y^2}{4} = 1$$ Desmos Link for More Detail$$\frac{x^2}{100} + \frac{y^2}{4} = 1$$

### Translations of Ellipses

Horizontal and vertical translations can be used to graph ellipses that are not centered at the origin. $$1) \frac{x^2}{9} + \frac{y^2}{4} = 1$$ $$2) \frac{(x - 2)^2}{9} + \frac{(y - 3)^2}{4} = 1$$ When we compare #1 to #2, we can see the graphs have the same shape and size, however, the graph of the first equation is centered at the origin, whereas, the graph of the second equation is centered at (2, 3). In other words, all the points have shifted 2 units right and 3 units up. Desmos Link for More Detail$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$

$$\frac{(x - 2)^2}{9} + \frac{(y - 3)^2}{4} = 1$$

If h and k are positive real numbers (h > 0 and k > 0):

Replacement | Shift |
---|---|

x is replaced with x - h | Shift the graph h units right |

x is replaced with x + h | Shift the graph h units left |

y is replaced with y - k | Shift the graph k units up |

y is replaced with y + k | Shift the graph k units down |

Example #3: Find the center, vertices, covertices, foci, and equation in standard form, and then sketch the graph. $$49x^2 + 9y^2 - 196x - 245 = 0$$ Group the x-terms together. $$49x^2 - 196x + 9y^2 - 245 = 0$$ Factor out the coefficient of x

^{2}: $$49(x^2 - 4x) + 9y^2 - 245 = 0$$ Complete the square: $$49(x^2 - 4x + 4 - 4) + 9y^2 - 245 = 0$$ Group the perfect square trinomial: $$49(x^2 - 4x + 4) - (49)(4) + 9y^2 - 245 = 0$$ $$49(x^2 - 4x + 4) - 196 + 9y^2 - 245 = 0$$ Simplify and factor: $$49(x - 2)^2 + 9y^2 - 441 = 0$$ Add 441 to each side: $$49(x - 2)^2 + 9y^2 = 441$$ Divide both sides by 441: $$\require{cancel}\frac{\cancel{49}(x - 2)^2}{9\cancel{441}} + \frac{\cancel{9}y^2}{49\cancel{441}} = \frac{1\cancel{441}}{\cancel{441}}$$ $$\frac{(x - 2)^2}{9} + \frac{y^2}{49} = 1$$ Since 49 > 9, the major axis will be vertical with a^{2}= 49 and b^{2}= 9. Compare to our formula: $$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$$ The replacement of x with (x - 2) gives us a shift 2 units right. $$\frac{(x - h)^2}{b^2} + \frac{y^2}{a^2} = 1$$ $$h = 2, b^2 = 9, a^2 = 49$$ $$\frac{(x - 2)^2}{9} + \frac{y^2}{49} = 1$$ Center: $$(2, 0)$$ To find the vertices, recall that a is the distance from the center to each vertex. $$a = \sqrt{49} = 7$$ This means we would travel from the center 7 units up and 7 units down to get to the vertices.Vertices: $$(2, -7), (2, 7)$$ To get the foci, recall that c is the distance from the center to each focus. $$c^2 = a^2 - b^2$$ $$a^2 = 49, b^2 = 9$$ Plug in: $$c^2 = 49 - 9 = 40$$ $$c = \sqrt{40} = 2\sqrt{10}$$ Foci: $$\left(2, 2\sqrt{10}\right), \left(2, -2\sqrt{10}\right)$$ To get the covertices, recall that b is the distance from the center to each covertex. $$b^2 = 9$$ $$b = 3$$ $$2 - 3 = -1, 2 + 3 = 5$$ Covertices: $$(-1, 0), (5, 0)$$ Desmos Link for More Detail

$$\frac{(x - 2)^2}{9} + \frac{y^2}{49} = 1$$

### Standard Forms of Equations of Ellipses Centered at (h, k)

Using a table to organize the formulas will speed up your work when the center is not at the origin. It is helpful to remember that a is the distance from the center to each vertex, b is the distance from the center to each covertex, and then c is the distance from the center to each focus.Equation | Center | Major Axis | Vertices | Covertices | Foci |
---|---|---|---|---|---|

$$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$$ | $$(h, k)$$ | Horizontal | $$(h - a,k)$$$$(h + a, k)$$ | $$(h, k - b)$$$$(h, k + b)$$ | $$(h - c, k)$$$$(h + c, k)$$ |

$$\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1$$ | $$(h, k)$$ | Vertical | $$(h,k - a)$$$$(h, k + a)$$ | $$(h - b, k)$$$$(h + b, k)$$ | $$(h, k - c)$$$$(h, k + c)$$ |

Example #4: Find the standard form of the ellipse and sketch the graph.

Vertices: $$(-10, 8), (-10, -18)$$ Foci: $$(-10, 7), (-10, -17)$$ Recall that the vertices lie on the major axis. Since we have the same x-coordinate of -10 in each case and different y-coordinates (8 and -18), we can conclude that the major axis is vertical. The center is the midpoint of the major axis. $$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ Plug into the midpoint formula: $$M = \left(\frac{-10 + (-10)}{2}, \frac{8 + (-18)}{2}\right) = (-10, -5)$$ Center: $$(-10, -5)$$ Let's plug in what we know so far. $$\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1$$ $$\frac{(x + 10)^2}{b^2} + \frac{(y + 5)^2}{a^2} = 1$$ a is the distance from the center to each vertex. $$a = 8 - (-5) = 8 + 5 = 13$$ $$a^2 = 13^2 = 169$$ $$\frac{(x + 10)^2}{b^2} + \frac{(y + 5)^2}{169} = 1$$ We can get b

^{2}from our formula: $$b^2 = a^2 - c^2$$ c is the distance from the center to each focus. $$c = 7 - (-5) = 7 + 5 = 12$$ $$c^2 = 12^2 = 144$$ $$b^2 = 169 - 144 = 25$$ $$\frac{(x + 10)^2}{25} + \frac{(y + 5)^2}{169} = 1$$ Covertices: $$(-5, -5), (-15, -5)$$ Desmos Link for More Detail$$\frac{(x + 10)^2}{25} + \frac{(y + 5)^2}{169} = 1$$

#### Skills Check:

Example #1

Find the standard form of the ellipse.

Vertices: $$(8, 0), (-8, 0)$$ Foci: $$\left(\sqrt{39}, 0\right), \left(-\sqrt{39}, 0\right)$$

Please choose the best answer.

A

$$\frac{x^2}{64} + \frac{y^2}{25} = 1$$

B

$$\frac{x^2}{25} + \frac{y^2}{64} = 1$$

C

$$\frac{x^2}{100} + \frac{y^2}{169} = 1$$

D

$$\frac{(x - 1)^2}{25} + \frac{y^2}{144} = 1$$

E

$$\frac{x^2}{64} + \frac{(y - 1)^2}{49} = 1$$

Example #2

Find the standard form of the ellipse.

Covertices: $$(19, 6), (-1, 6)$$ Foci: $$\left(9, 6 + 2\sqrt{11}\right), \left(9, 6 - 2\sqrt{11}\right)$$

Please choose the best answer.

A

$$\frac{(x - 9)^2}{36} + \frac{(y - 6)^2}{144} = 1$$

B

$$\frac{(x - 9)^2}{36} + \frac{(y + 6)^2}{144} = 1$$

C

$$\frac{(x - 12)^2}{100} + \frac{(y + 3)^2}{225} = 1$$

D

$$\frac{(x - 9)^2}{144} + \frac{(y - 6)^2}{100} = 1$$

E

$$\frac{(x - 9)^2}{100} + \frac{(y - 6)^2}{144} = 1$$

Example #3

Find the standard form of the ellipse. $$36x^2 + y^2 + 4y - 32 = 0$$

Please choose the best answer.

A

$$\frac{x^2}{36} + \frac{y^2}{25} = 1$$

B

$$\frac{(x - 1)^2}{144} + \frac{(y - 2)^2}{36} = 1$$

C

$$x^2 + \frac{(y + 2)^2}{36} = 1$$

D

$$x^2 + \frac{(y - 2)^2}{16} = 1$$

E

$$\frac{x^2}{36} + y^2 = 1$$

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