Mastery Check Solutions

1. Determine which of the given equations is a hyperbola. Explain. $J: 4x^2+2y^2=32x+32y–2y^2 K: y^2=4x^2+32x+100 L:x^2+2y^2=x^2+64y+32x$

$$\begin{gathered} J: 4x^2+0xy+4y^2–32x–32y+0=0 \\ A=4, C=4 \\ K:4x^2+0xy–y^2+32x+0y+100=0 \\ A=4, C=–1 \\ L:= 0x^2+0xy+2y^2–32x–64y+0=0 \\ A=0, C=2 \end{gathered}$$

Equation K is a hyperbola because A and C are non-zero and have opposite signs.

2. Write the equation of a hyperbola from part A in standard form. Name the center and equation for the asymptotes.

$$\begin{gathered} 4x^2+32x–y^2=–100 \\ \left(4x^2+32x\right)–y^2=–100 \\ 4\left(x^2+8x+{\left(\boxed{\frac{8}{2}}\right)}^2\right)–y^2=–100+4{\left(\boxed{\frac{8}{2}}\right)}^2 \\ 4{\left(x+4\right)}^2–y^2=–100+4{\left(4\right)}^2 \\ 4{\left(x+4\right)}^2–y^2=–100+4\left(16\right) \\ \frac{4{\left(x+4\right)}^2}{–36}–\frac{y^2}{–36}=\frac{–36}{–36} \\ \frac{{\left(x+4\right)}^2}{–9}+\frac{y^2}{36}=1 \\ \frac{y^2}{36}–\frac{{\left(x+4\right)}^2}{9}=1 \end{gathered}$$

$$\begin{gathered} \mathrm{Center} \left(–4, 0\right) \\ a=3, b=6 \\ m=\pm \frac{b}{a}=\pm \frac{6}{3} \\ \mathrm{A}\mathrm{symptotes}: y=\pm 2\left(x+4\right) \end{gathered}$$