Graphing Ellipses Practice

The strategy

If we have the center, the vertical radius, and the horizontal radius of our ellipse, we can find its equation by substituting these values into the standard equation of the ellipse.

This equation represents an ellipse with center $(h, k)$, a horizontal radius of ‍$a$, and a vertical radius of ‍$b$.

‍$\dfrac{(x - h)^2}{ a^2} + \dfrac{(y - k)^2}{b^2} = 1$

Writing the equation of the ellipse

From the graph, we can see that the ellipse is centered at‍ $({2},{0})$, has a horizontal radius of ‍‍$7$, and a vertical radius of ‍‍$6$.

Therefore, we can write the standard equation of our ellipse as follows.

‍$\dfrac{(x - ({2}))^2}{ 7 ^2} + \dfrac{(y - ({0}))^2}{6^2} = 1$

We can simplify this equation by evaluating the squares.

‍$\dfrac{(x-{2}) ^2}{ {49}} + \dfrac{y^2}{{36}} = 1$

Summary

The equation of the graphed ellipse is given below.

‍$\dfrac{(x-2)^2}{49} + \dfrac{y^2}{36} = 1$

The strategy

If we have the center, the vertical radius, and the horizontal radius of our ellipse, we can find its equation by substituting these values into the standard equation of the ellipse.

This equation represents an ellipse with center $(h, k)$, a horizontal radius of ‍$a$, and a vertical radius of ‍$b$.

‍$\dfrac{(x - h)^2}{ a^2} + \dfrac{(y - k)^2}{b^2} = 1$

Writing the equation of the ellipse

From the graph, we can see that the ellipse is centered at‍ $({-3},{-4})$, has a horizontal radius of ‍‍$4$, and a vertical radius of ‍‍$5$.

Therefore, we can write the standard equation of our ellipse as follows.

‍$\dfrac{(x - ({{-3}}))^2}{ 4 ^2} + \dfrac{(y - ({-4}))^2}{5 ^2} = 1$

We can simplify this equation by evaluating the squares.

‍$\dfrac{(x + {3})^2}{ {16}} + \dfrac{(y + 4)^2}{ {25}} = 1$

Summary

The equation of the graphed ellipse is given below.

‍$\dfrac{(x +3)^2}{16} + \dfrac{(y +4)^2}{25} = 1$

The strategy

The standard equation of an ellipse with center $(h, k)$, a horizontal radius of ‍$a$, and a vertical radius of ‍$b$.

‍$\dfrac{(x - h)^2}{ a^2} + \dfrac{(y - k)^2}{b^2} = 1$

We can rewrite the equation of our ellipse in this form to find its center and radii. Then, we can find the graph that correctly represents our ellipse.

Rewriting the equation

$\begin{aligned} \dfrac{(x+5)^2}{9}+\dfrac{(y-3)^2}{25}&=1 \\\\ \dfrac{(x - ({-5}))^2}{ 3^2} + \dfrac{(y - ({3}))^2}{ 5^2} &= 1\end{aligned}$

Therefore, our ellipse is centered at $({-5}, {3})$, has a horizontal radius of ‍$3$ units, and a vertical radius of ‍$5$ units.

Selecting the correct graph

Only graph C contains the ellipse with the center ‍\(\), a horizontal radius of ‍$3$ units, and a vertical radius of ‍$5$ units.

Summary

Graph C contains the ellipse represented by the given equation.

The strategy

If we have the center, the vertical radius, and the horizontal radius of our ellipse, we can find its equation by substituting these values into the standard equation of the ellipse.

This equation represents an ellipse with center $(h, k)$, a horizontal radius of ‍$a$, and a vertical radius of ‍$b$.

‍$\dfrac{(x - h)^2}{ a^2} + \dfrac{(y - k)^2}{b^2} = 1$

Writing the equation of the ellipse

From the graph, we can see that the ellipse is centered at‍ $({2},{1})$, has a horizontal radius of ‍‍$4$, and a vertical radius of ‍‍$6$.

Therefore, we can write the standard equation of our ellipse as follows.

‍$\dfrac{(x - (2))^2}{ 4^2} + \dfrac{(y - (1))^2}{ 6^2} = 1$

We can simplify this equation by evaluating the squares.

‍$\dfrac{(x - {2})^2}{ {16}} + \dfrac{(y - 1)^2}{ {36}} = 1$

Summary

The equation of the graphed ellipse is given below.

‍$\dfrac{(x -2)^2}{16} + \dfrac{(y - 1)^2}{36} = 1$