Test 15 (Lessons 29–30): Ellipses and Hyperbolas Solutions

For problems 1–2, write the equation in standard form.

An ellipse standard form

center (4, –1)
a = 3, b = 5

$\frac{{(x - 4)}^{2}}{9} + \frac{{(y + 1)}^{2}}{49} = 1$

A hyperbola

center (–5, 3)
a = 5, b = 8

$\frac{{(x + 5)}^{2}}{25} - \frac{{(y - 3)}^{2}}{64} = 1$

Write and graph the equation of a horizontal ellipse translated 2 spaces down from the center with a domain of $[- 11, 11]$ and a range of $[- 8, 4] .$

$(0, 0) \to (0, 0 - 2)$
center (0, –2)

major: $\frac{22}{2} = 11$
$11^{2} = 121$

minor: $\frac{12}{2} = 6$
$6^{2} = 36$

$\frac{x^{2}}{121} + \frac{{(y + 2)}^{2}}{36} = 1$

Describe how to determine if an ellipse will be horizontal or vertical when only given the equation in standard form.

Sample: The values of a and b determine the major and minor axes. When $a > b,$ the ellipse has a horizontal major axis. When $b > a,$ the ellipse has a vertical major axis.

Explain how to find the asymptotes when given the equation of a hyperbola in standard form.

Sample: The center of the hyperbola and the slope are needed to write the equations of the asymptotes in point-slope form. The equation will be: $y - k = \pm \frac{b}{a} (x - h)$

Graph.
$\frac{{(y - 2.5)}^{2}}{4} - \frac{{(x + 3)}^{2}}{100} = 1$

(–3, 2.5)

b = 4
a = 10

$m = \pm \frac{b}{a} = \pm \frac{4}{10} m = \pm \frac{2}{5}$

For problems 7–8, name the conic section from the given equation. Explain your reasoning.

$4 x^{2} + 10 y = 25 y^{2} + 5 x + 20$

$4 x^{2} - 25 y^{2} - 5 x + 10 y + 20 = 0 4 x^{2} + 0 x y - 25 y^{2} - 5 x + 10 y + 20 = 0 A = 4, B = 0, C = - 25$

The equation represents a hyperbola because both x and y have squared terms and the values of A and C are non-zero with different signs.

$2 y^{2} + 3 y = 4 x + 7$

$2 y^{2} - 4 x + 3 y - 7 = 0 0 x^{2} + 0 x y + 2 y^{2} - 4 x + 3 y - 7 = 0$

The equation represents a parabola because $A = 0$ , and there is only one term raised to the second power. (Optional: The parabola opens to the right.)

Write the equation in standard form. Name the type of conic and its center.

$4 x^{2} + 25 y^{2} + 50 y = 66 + 12 x$

$4 x^{2} - 12 x + 25 y^{2} + 50 y = 66 4 (x^{2} - 3 x) + 25 (y^{2} + 2 y) = 66 4 (x^{2} - 3 x + {(\frac{- 3}{2})}^{2}) + 25 (y^{2} + 2 y + {(\frac{2}{2})}^{2}) = 66 + 4 ({(\frac{- 3}{2})}^{2}) + 25 ({(\frac{2}{2})}^{2}) 4 {(x - \frac{3}{2})}^{2} + 25 {(y + 1)}^{2} = 66 + 4 (\frac{9}{4}) + 25 (1) 4 {(x - 1.5)}^{2} + 25 {(y + 1)}^{2} = 66 + 9 + 25 \frac{4 {(x - 1.5)}^{2}}{100} + \frac{25 {(y + 1)}^{2}}{100} = \frac{100}{100}$

$\frac{{(x - 1.5)}^{2}}{25} + \frac{{(y + 1)}^{2}}{4} = 1$

This is an ellipse with a center at $(1.5, - 1) .$

A group of students was told that the equation $5 y^{2} + 12 x + 10 y = 3 x^{2} + 22$ would result in a hyperbola, but when they solved it, their result was an ellipse. Determine the correct equation and explain your reasoning.

Student work (contains error)
$3 x^{2} + 12 x + 5 y^{2} + 10 y = 22 3 (x^{2} + 4 x + (2^{2})) + 5 (y^{2} + 2 y + (1^{2})) = 22 + 3 (2^{2}) + 5 (1^{2}) 3 {(x + 2)}^{2} + 5 {(y + 1)}^{2} = 22 + 12 + 5 3 {(x + 2)}^{2} + 5 {(y + 1)}^{2} = 39 \frac{3 {(x + 2)}^{2}}{39} + \frac{5 {(y + 1)}^{2}}{39} = 1$

$- 3 x^{2} + 12 x + 5 y^{2} + 10 y = 22 - 3 (x^{2} - 4 x + {(\frac{- 4}{2})}^{2}) + 5 (y^{2} + 2 y + {(\frac{2}{2})}^{2}) = 22 - 3 {(\frac{- 4}{2})}^{2} + 5 {(\frac{2}{2})}^{2} - 3 {(x - 2)}^{2} + 5 {(y + 1)}^{2} = 22 - 12 + 5 \frac{- 3 {(x - 2)}^{2}}{15} + \frac{5 {(y + 1)}^{2}}{15} = \frac{15}{15} \frac{{(x - 2)}^{2}}{- 5} + \frac{{(y + 1)}^{2}}{3} = 1 \frac{{(y + 1)}^{2}}{3} - \frac{{(x - 2)}^{2}}{5} = 1$

The group of students did not subtract $3 x^{2}$ from both sides. In order to have a hyperbola, $x^{2}$ and $y^{2}$ need to have opposite signs.