Practice Solutions

1. $\frac{{\left(x+5\right)}^2}{289}+\frac{{\left(y–6\right)}^2}{64}=1$

$$\begin{gathered} a^2=289 \\ {b}^2=64 \\ \begin{array}{c}\begin{array}{rcl}{a}^2–b^2& =& c^2\\ 289–64& =& c^2\\ \sqrt{c^2}& =& \sqrt{225}\\ c& =& 15\end{array}\end{array} \\ \left(h, k\right): \left(–5, 6\right) \\ \left(h\pm c, k\right): \left(–5\pm 15, 6\right) \end{gathered}$$

Horizontal ellipse, foci: $\left(–20, 6\right), \left(10, 6\right)$

2. $\frac{x^2}{36}–\frac{{\left(y+2\right)}^2}{64}=1$

$$\begin{gathered} a^2=36 \\ {b}^2=64 \\ \begin{array}{c}\begin{array}{rcl}{a}^2+b^2& =& c^2\\ 36+64& =& c^2\\ \sqrt{c^2}& =& \sqrt{100}\\ c& =& 10\end{array}\end{array} \\ \left(h, k\right): \left(0, –2\right) \\ \left(h, k\pm c\right)=\left(0, –2\pm 10\right) \end{gathered}$$

Vertical hyperbola, foci: $\left(0, –12\right), \left(0, 8\right)$

3. $\frac{{\left(x+1\right)}^2}{144}+\frac{{\left(y–8\right)}^2}{169}=1$

$$\begin{gathered} a^2=169 \\ {b}^2=144 \\ \begin{array}{c}\begin{array}{rcl}{a}^2–b^2& =& c^2\\ 169–144& =& c^2\\ \sqrt{c^2}& =& \sqrt{25}\\ c& =& 5\end{array}\end{array} \\ \left(h, k\right): \left(–1, 8\right) \\ \left(h, k\pm c\right): \left(–1, 8\pm 5\right) \end{gathered}$$

Vertical ellipse, foci: $\left(–1, 3\right), \left(–1, 13\right)$

4. $\frac{x^2}{289}+\frac{y^2}{64}=1$

$$\begin{gathered} a^2=289 \\ {b}^2=64 \\ \begin{array}{c}\begin{array}{rcl}{a}^2–b^2& =& c^2\\ 289–64& =& c^2\\ \sqrt{c^2}& =& \sqrt{225}\\ c& =& 15\end{array}\end{array} \\ \left(h, k\right): \left(0, 0\right) \\ \left(h\pm c, k\right): \left(0\pm 15, 0\right) \end{gathered}$$

Horizontal ellipse, foci: $\left(–15, 0\right), \left(15, 0\right)$

5. $\frac{y^2}{225}–\frac{x^2}{64}=1$

$$\begin{gathered} a^2=225 \\ {b}^2=64 \\ \begin{array}{c}\begin{array}{rcl}{a}^2+b^2& =& c^2\\ 225+64& =& c^2\\ \sqrt{c^2}& =& \sqrt{289}\\ c& =& 17\end{array}\end{array} \\ \left(h, k\right): \left(0, 0\right) \\ \left(h, k\pm c\right): \left(0, 0\pm 17\right) \end{gathered}$$

Vertical hyperbola, foci: $\left(0, –17\right), \left(0, 17\right)$

6. $\frac{{\left(y–17\right)}^2}{49}–\frac{{\left(x+15\right)}^2}{576}=1$

$$\begin{gathered} a^2=49 \\ {b}^2=576 \\ \begin{array}{c}\begin{array}{rcl}{a}^2+b^2& =& c^2\\ 49+576& =& c^2\\ \sqrt{c^2}& =& \sqrt{625}\\ c& =& 25\end{array}\end{array} \\ \left(h, k\right): \left(17, –15\right) \\ \left(h, k\pm c\right): \left(17, –15\pm 25\right) \end{gathered}$$

Vertical hyperbola, foci: $\left(17, –40\right), \left(17, 10\right)$

7. A horizontal ellipse with vertices at $\left(15, 0\right)$ and $\left(–15, 0\right)$ and focal length of 12 units.

$$\begin{gathered} \mathrm{Center}: \left(\frac{15+\left(–15\right)}{2}, \frac{0+0 }{2}\right)=\left(0, 0\right) \\ c=12 \\ \begin{array}{lcl}\mathrm{Find} a:& & \mathrm{Find} b:\\ \left(0, 0\right), \left(15, 0\right)& & a^2–b^2=c^2\\ a=\sqrt{{\left(15–0\right)}^2+{\left(0–0\right)}^2}& & \begin{array}{rcl}{15}^2–b^2& =& {12}^2\end{array}\\ a=15& & \begin{array}{rcl}225–b^2& =& 144\end{array}\\ & & \begin{array}{rcl} –b^2& =& –81\end{array}\\ & & \begin{array}{rcl} b& =& 9\end{array}\end{array} \end{gathered}$$

$\frac{x^2}{225}+\frac{y^2}{81}=1$

8. A hyperbola with a horizontal transverse axis of 10 and foci at $\left(0, 0\right)$ and $\left(26, 0\right)$.

$$\begin{gathered} \mathrm{Center}: \left(\frac{0+26}{2}, \frac{0+0 }{2}\right)=\left(13, 0\right) \\ \mathrm{Find} c: \\ \left(13, 0\right), \left(0, 0\right) \\ c=\sqrt{{\left(0–13\right)}^2+{\left(0–0\right)}^2} \\ c=13 \\ \begin{array}{lcl}\mathrm{Find} a:& & \mathrm{Find} b:\\ 2a=10& & a^2+b^2=c^2\\ a=5& & 5^2+b^2={13}^2\\ & & 25+b^2=169\\ & & b^2=144\\ & & b=12\end{array} \end{gathered}$$

$\frac{{\left(x–13\right)}^2}{25}–\frac{y^2}{144}=1$

9. A horizontal ellipse with vertices at $\left(17, 0\right)$ and $\left(–17, 0\right)$ and a focal length of 8 units.

$$\begin{gathered} \mathrm{Center}: \left(\frac{17+\left(–17\right)}{2}, \frac{0+0 }{2}\right)=\left(0, 0\right) \\ c=8 \\ \begin{array}{lcl}\mathrm{Find} a:& & \mathrm{Find} b:\\ \left(0, 0\right), \left(17, 0\right)& & a^2–b^2=c^2\\ a=\sqrt{{\left(17–0\right)}^2+{\left(0–0\right)}^2}& & {17}^2–b^2=8^2\\ a=17& & 289–b^2=64 \\ –b^2=–225 \\ b=15\end{array} \end{gathered}$$

$\frac{x^2}{289}+\frac{y^2}{225}=1$

10. An ellipse with focal points at $\left(–2, 5\right)$ and $\left(–2, –11\right)$ and co-vertices at $(4, –3)$ and $(–8, –3)$.

$$\begin{gathered} \mathrm{Center}: \left(\frac{–2+\left(–2\right)}{2}, \frac{5+\left(–11\right) }{2}\right)=\left(–2, –3\right) \\ \mathrm{Find} c: \\ \left(–2, –3\right), \left(–2, 5\right) \\ c=\sqrt{{\left(–2–\left(–2\right)\right)}^2+{\left(–3–5\right)}^2}=8 \\ \begin{array}{lcl}\mathrm{Find} b:& & \mathrm{Find} a:\\ \left(–2, –3\right), \left(4, –3\right)& & a^2–b^2=c^2\\ b=\sqrt{{\left(–2–4\right)}^2+{\left(–3–\left(–3\right)\right)}^2}& & a^2–6^2=8^2\\ b=6& & a^2–36=64\\ & & a^2=100\\ & & a=10\end{array} \end{gathered}$$

$\frac{{\left(x+2\right)}^2}{36}+\frac{{\left(y+3\right)}^2}{100}=1$

11. A hyperbola with a vertical transverse axis of 48 and foci at $(2, 1)$ and $(2, 51)$.

$$\begin{gathered} \mathrm{Center}: \left(\frac{2+2}{2}, \frac{ 1+51}{2}\right)=\left(2, 26\right) \\ \mathrm{Find} c: \\ \left(2, 26\right), \left(2, 1\right) \\ c=\sqrt{{\left(2–2\right)}^2+{\left(26–1\right)}^2} \\ c=25 \\ \begin{array}{ccl}\mathrm{Find} a:& & \mathrm{Find} b:\\ 2a=48& & a^2+b^2=c^2\\ a=24& & {24}^2+b^2={25}^2\\ & & 576+b^2=625\\ & & b^2=49\\ & & b=7\end{array} \end{gathered}$$

$\frac{{\left(y–26\right)}^2}{576}–\frac{{\left(x–2\right)}^2}{49}=1$

12. An ellipse with focal points at $(2, 8)$ and $(14, 8)$ and vertices at $(0, 8)$ and $(16, 8)$.

$$\begin{gathered} \mathrm{Center}: \left(\frac{2+14}{2}, \frac{8+8 }{2}\right)=\left(8, 8\right) \\ \mathrm{Find} c: \\ \left(8, 8\right), \left(2, 8\right) \\ c=\sqrt{{\left(2–8\right)}^2+{\left(8–8\right)}^2} \\ c=6 \\ \begin{array}{lcl}\mathrm{Find} a:& & \mathrm{Find} b:\\ \left(8, 8\right), \left(0, 8\right)& & a^2–b^2=c^2\\ a=\sqrt{{\left(0–8\right)}^2+{\left(8–8\right)}^2}& & 8^2–b^2=6^2\\ a=8& & 64–b^2=36\\ & & –b^2=–28\\ & & b=\sqrt{28}\end{array} \end{gathered}$$

$\frac{{\left(x–8\right)}^2}{64}+\frac{{\left(y–8\right)}^2}{28}=1$

13. A hyperbola with a horizontal transverse length of 12 and foci at $(–7, 0)$ and $(7, 0)$.

$$\begin{gathered} \mathrm{Center}: \left(\frac{–7+7}{2}, \frac{ 0+0}{2}\right)=\left(0, 0\right) \\ \mathrm{Find} c: \\ \left(0, 0\right), \left(–7, 0\right) \\ c=\sqrt{{\left(–7–0\right)}^2+{\left(0–0\right)}^2} \\ c=7 \\ \begin{array}{lcl}\mathrm{Find} a:& & \mathrm{Find} b:\\ 2a=12& & a^2+b^2=c^2\\ a=6& & 6^2+b^2=7^2\\ & & 36+b^2=49\\ & & b^2=15\\ & & b=\sqrt{15}\end{array} \end{gathered}$$

$\frac{x^2}{36}–\frac{y^2}{15}=1$

14. A hyperbola with vertices at $(5, 2)$ and $(1, 2)$ and focal points at $(7, 2)$ and $(–1, 2)$.

$$\begin{gathered} \mathrm{Center}: \left(\frac{5+1}{2}, \frac{2+2}{2}\right)=\left(3, 2\right) \\ \mathrm{Find} c: \\ \left(3, 2\right), \left(7, 2\right) \\ c=\sqrt{{\left(7–3\right)}^2+{\left(2–2\right)}^2} \\ c=4 \\ \begin{array}{lcl}\mathrm{Find} a:& & \mathrm{Find} b:\\ \left(3, 2\right), \left(5, 2\right)& & a^2+b^2=c^2\\ a=\sqrt{{\left(5–3\right)}^2+{\left(2–2\right)}^2}& & 2^2+b^2=4^2\\ a=2& & 4+b^2=16\\ & & b^2=12\\ & & b=2\sqrt{3}\end{array} \end{gathered}$$

$\frac{{\left(x–3\right)}^2}{16}–\frac{{\left(y–2\right)}^2}{12}=1$