Practice 1 Solutions

Write the equation of an ellipse in standard form that satisfies each set of conditions.

The center of an ellipse is $(- 6.3, 19.2)$ and has a major horizontal radius of $a = \sqrt{61}$ units and a minor axis length of $b = 8$ units.

$Center : (- 6.3, 19.2) a = \sqrt{61} a^{2} = 61 2 b = 8 b = 4 b^{2} = 16$

$\frac{{(x + 6.3)}^{2}}{61} + \frac{{(y - 19.2)}^{2}}{16} = 1$

Note

Q: How do you determine the value of a or b when given the length of an axis?

A: Divide the length of the axis by two and then square the value.

Write the equation of the ellipse from the graph and label the center, vertices, and co-vertices.

$Center : (4, 0) 2 a = 10 a = 5 a^{2} = 25 2 b = 14 b = 7 b^{2} = 49$

$\frac{{(x - 4)}^{2}}{25} + \frac{y^{2}}{49} = 1$

Find the equation of the ellipse with vertices: $(- 10, 5)$ , $(0, 5)$ , $(- 5, 16)$ , $(- 5, - 6)$

$Center : (\frac{- 10 + 0}{2}, \frac{5 + 5}{2}) = (- 5, 5) d = \sqrt{{(- 10 - 0)}^{2} + {(5 - 5)}^{2}} = 10 2 a = 10 a = 5 d = \sqrt{{(- 5 - (- 5))}^{2} + {(16 - (- 6))}^{2}} = 22 2 b = 22 b = 11$

$\frac{{(x + 5)}^{2}}{25} + \frac{{(y - 5)}^{2}}{121} = 1$

Note

Q: What formulas do you need to find the equation of the ellipse when given the vertices and co-vertices?

A: First, you need the midpoint formula to find the center, and then you need the distance formula to calculate the distance of the axes.

Write the equation of the ellipse tangent to: $x = 5, x = - 2, y = 5, y = - 6$

$Center : (\frac{5 + - 2}{2}, \frac{5 + - 6}{2}) = (1.5, - 0.5) 2 a = 7 a = 3.5 a^{2} = 12.25 2 b = 11 b = 5.5 b^{2} = 30.25$

$\frac{{(x - 1.5)}^{2}}{12.25} + \frac{{(y + 0.5)}^{2}}{30.25} = 1$

Write the equation of the ellipse when the given graph is translated 3 right and 1 down.

$Center : (2 + 3, - 1 - 1) = (5, - 2) d = \sqrt{{(- 8 - 12)}^{2} + {(- 1 - (- 1))}^{2}} = 20 2 a = 20 a = 10 a^{2} = 100 d = \sqrt{{(2 - 2)}^{2} + {(5 - (- 7))}^{2}} = 12 2 b = 12 b = 6 b^{2} = 36$

$\frac{{(x - 5)}^{2}}{100} + \frac{{(y + 2)}^{2}}{36} = 1$

Write the equation of the ellipse when it is translated 2 to the left and 6 down, and the major axis and minor axis are doubled.
$x^{2} + \frac{y^{2}}{4} = 1$

$Center : (0 - 2, 0 - 6) = (- 2, - 6) a = 1 2 a = 2 Transformed 2 (2 a) = 2 (2) = 4 Transformed a = 2 b = 2 2 b = 4 Transformed 2 (2 b) = 2 (4) = 8 Transformed b = 4$

$\frac{{(x + 2)}^{2}}{4} + \frac{{(y + 6)}^{2}}{16} = 1$

The National Statuary Hall in the U.S. Capitol building is an elliptical room measuring 46 feet by 97 feet. With a center at the origin, write the equation to represent the footprint of the room.

$2 a = 46 a = 23 a^{2} = 529 2 b = 97 b = 48.5 b^{2} = 2,352 . 25$

$\frac{x^{2}}{529} + \frac{y^{2}}{2,352 . 25} = 1$

Note

It would also be correct if $a = 48.5, b = 23$ , making the equation $\frac{x^{2}}{2,352 . 25} + \frac{y^{2}}{529} = 1$ . The More to Explore for this lesson builds on the concept of Statuary Hall, including why it is nicknamed “The Whisper Gallery.”

A new arena is being built for music and sports performances. The footprint of the building will be elliptical with the length of the axes 288 meters and 432 meters. Write the equation of a model of the elliptical arena centered around the origin that will be 144th the size of the original.

$\begin{matrix} \frac{288}{144} = 2 & \frac{432}{144} = 3 \\ Center & (0, 0) \end{matrix}$

$\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$

Note

You will need to divide each axis by the scale factor to write the equation of the model.

Write the equation in standard form.

$4 x^{2} - 24 x + 16 y^{2} = 28$

$4 (x^{2} - 6 x + {(\frac{- 6}{2})}^{2}) + 16 y^{2} = 28 + 4 {(\frac{- 6}{2})}^{2} 4 {(x - 3)}^{2} + 16 y^{2} = 28 + 4 (9) \frac{4 {(x - 3)}^{2}}{64} + \frac{16 y^{2}}{64} = \frac{64}{64}$

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

Note

Q: What process is needed to rewrite an equation in standard form?

A: Completing the square

$x^{2} + 9 y^{2} + 4 x - 18 y = 23$

$x^{2} + 4 x + {(\frac{4}{2})}^{2} + 9 (y^{2} - 2 y + {(- \frac{2}{2})}^{2}) = 23 + {(\frac{4}{2})}^{2} + 9 {(- \frac{2}{2})}^{2} {(x + 2)}^{2} + 9 {(y - 1)}^{2} = 23 + 4 + 9 \frac{{(x + 2)}^{2}}{36} + \frac{9 {(y - 1)}^{2}}{36} = \frac{36}{36}$

$\frac{{(x + 2)}^{2}}{36} + \frac{{(y - 1)}^{2}}{4} = 1$

Explain the relationship between a circle and an ellipse.

A circle is a special type of ellipse where $a = b$ .

What points on an ellipse determine the domain and range?

$Domain : [h \pm a] Range : [k \pm b]$

Note

You can also write the domain and range rather than using interval notation.

Graph.

Graph: $\frac{{(x - 2)}^{2}}{9} + \frac{{(y + 2)}^{2}}{4} = 1$
Label the center, vertices, and co-vertices.

$Center : (2, - 2) a^{2} = 9 a = 3 b^{2} = 4 b = 2$

Graph: $\frac{x^{2}}{36} + \frac{y^{2}}{64} = 1$
Label the center, vertices, and co-vertices.

$Center : (0, 0) a^{2} = 36 a = 6 b^{2} = 64 b = 8$

The major vertical axis of an ellipse is 13 units with co-vertices $(2, 7)$ and $(7, 7)$ . Determine the equation, then graph.

$Center : (\frac{2 + 7}{2}, \frac{7 + 7}{2}) = (4.5, 7) 2 b = 13 b = 6.5 b^{2} = 42.25 2 a = 5 a = 2.5 a^{2} = 6.25$

$\frac{{(x - 4.5)}^{2}}{6.25} + \frac{{(y - 7)}^{2}}{42.25} = 1$

Graph the ellipse with the center $(3, 2)$ , with a major horizontal axis length of 24 units, and a minor axis length of 18 units. Label the center, vertices, and co-vertices.

$Center : (3, 2) 2 a = 24 a = 12 2 b = 18 b = 9$

Graph $\frac{{(x - 2.5)}^{2}}{49} + \frac{(y - 3.5)}{9} = 1$ if it was translated 4 units to the left and 3 units down. Label the center, vertices, and co-vertices.

$Center : (2.5 - 4, 3.5 - 3) = (- 1.5, 0.5) a^{2} = 49 a = 7 b^{2} = 9 b = 3$

Graph: $\frac{x^{2}}{12} + \frac{y^{2}}{30} = 1$
Label the center, vertices, and co-vertices.

$Center : (0, 0) a^{2} = 12 a = \sqrt{12} \approx 3.464 b^{2} = 30 b = \sqrt{30} \approx 5.477$

Note

When exact values are unknown, make your best estimate of points on a graph. You can also use technology to check your graphs.