Practice 2 Solutions

5. Graph the hyperbola. Mark the center, vertices and co-vertices, and asymptotes of ${(x+1)}^2–\frac{y^2}{25}=1$. Then write the equation of the asymptotes.

$$\begin{gathered} \mathrm{C}\mathrm{enter}: \left(–1, 0\right) \\ a=1 \\ b=5 \\ \mathrm{S}\mathrm{lope} \mathrm{of} \mathrm{asymptotes}: \pm 5 \end{gathered}$$

$y=\pm 5\left(x+1\right)$

6. Graph the hyperbola. Mark the center, vertices and co-vertices, and asymptotes of $\frac{{(y–3)}^2}{25}–\frac{x^2}{4}=1$. Then write the equation of the asymptotes.

$$\begin{gathered} \mathrm{C}\mathrm{enter}: \left(0, 3\right) \\ a=2 \\ b=5 \\ \mathrm{S}\mathrm{lope} \mathrm{of} \mathrm{asymptotes}:\hspace{0.17em}\pm \frac{5}{2} \end{gathered}$$

$y–3=\pm \frac{5}{2}x$

7. Graph the horizontal hyperbola when the equation of the asymptotes is $y=\pm \frac{2}{3}\left(x–6\right)$. Then write the equation of the hyperbola.

$$\begin{gathered} a=3 \\ {a}^2=9 \\ b=2 \\ {b}^2=4 \\ \mathrm{C}\mathrm{enter}: \left(6,0\right) \end{gathered}$$

$\frac{{\left(x–6\right)}^2}{9}–\frac{y^2}{4}=1$

8. Write the equation and graph the hyperbola $6y^2–36y–6x^2+12x=102$.

$$\begin{gathered} 6\left(y^2–6y\right)–6\left(x^2–2x\right)=102 \\ 6\left(y^2–6y+\boxed{{\left(–\frac{6}{2}\right)}^2}\right)–6\left(x^2–2x+\boxed{{\left(–\frac{2}{2}\right)}^2}\right)=102+6\boxed{{\left(–\frac{6}{2}\right)}^2}–6\boxed{{\left(–\frac{2}{2}\right)}^2} \\ 6{\left(y–3\right)}^2–6{\left(x–1\right)}^2=102+6\left(9\right)–6\left(1\right) \\ 6{\left(y–3\right)}^2–6{\left(x–1\right)}^2=150 \end{gathered}$$

$\frac{{\left(y–3\right)}^2}{25}–\frac{{\left(x–1\right)}^2}{25}=1$

9. Find an equation of the vertical hyperbola when the equation of the asymptotes is $y+2=\pm \left(x+1\right)$.

$$\begin{gathered} \mathrm{C}\mathrm{enter}: \left(–1, –2\right) \\ \mathrm{S}\mathrm{lope} \mathrm{of} \mathrm{asymptote}: \pm 1 \\ a=1 \\ b=1 \end{gathered}$$

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

10. Write the equation of the hyperbola $8x^2–8x–8y^2–32y=158$.

$$\begin{gathered} 8\left(x^2–x\right)–8\left(y^2–4y\right)=158 \\ 8\left(x^2–x+\boxed{{\left(–\frac{1}{2}\right)}^2}\right)–8\left(y^2–4y+\boxed{{\left(–\frac{4}{2}\right)}^2}\right)=158+8\boxed{{\left(–\frac{1}{2}\right)}^2}–8\boxed{{\left(–\frac{4}{2}\right)}^2} \\ 8{\left(x–\frac{1}{2}\right)}^2–8{\left(y–2\right)}^2=158+8\left(\frac{1}{4}\right)–8\left(4\right) \\ 8{\left(x–\frac{1}{2}\right)}^2–8{\left(y–2\right)}^2=128 \end{gathered}$$

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

11. $2y=3x^2–6x–2$

$A=3, C=0$

B

12. $x^2–y^2=1$

A and C have different signs. 

A

13. $x^2+3y^2=12$

A and C have the same sign. $A\ne C$

C

14. $\frac{2{\left(x+4\right)}^2}{15}–4y=1$

$A=\frac{2}{15}, C=0$

Parabola

15. $\frac{6y^2}{13}–x^2=4$

A and C have different signs.

Hyperbola

16. $x^2+6y^2–14y=7y^2–10$

$x^2–y^2–14y=–10$
A and C have different signs.

Hyperbola

17. $–4x^2–8x+9=4y^2–10$

$4x^2+8x+4y^2=19$
$A=C$

Circle

18. $32x–16x^2+14y–7y^2=12$

$$\begin{gathered} –16x^2+32x–7y^2+14y=12 \\ 16x^2–32x+7y^2–14y=–12 \\ A\ne C. A\mathit{ }\mathrm{and} C\mathit{ }\mathrm{have} \mathrm{the} \mathrm{same} \mathrm{sign}. \end{gathered}$$

Ellipse