Hyperbola Practice

Does the equation represent horizontal hyperbola or a vertical hyperbola?

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

• horizontal

• vertical

What is the center of the graph of  $\frac{\left(y-1\right)^2}{16}-\frac{\left(x+1\right)^2}{4}=1$  

• (1, -1)

• (-1, 1)

• (-1, -1)

• (1, 1)

What are the vertices of the graph of  $\frac{\left(y-1\right)^2}{16}-\frac{\left(x+1\right)^2}{4}=1$  

• (-1, 5) and (-1, -3)

• (1, 3) and (1, -1)

• (5, 1) and (-3, 1)

• (1, 5) and (1, -3)

What are the foci of the graph of  $\frac{\left(y-1\right)^2}{16}-\frac{\left(x+1\right)^2}{4}=1$  

•  $\left(-1,\ 1\pm2\sqrt{5}\right)$  

•  $\left(1,\ 1\pm2\sqrt{5}\right)$  

•  $\left(1\pm2\sqrt{5},\ 1\right)$  

•  $\left(1,\ 1\pm2\sqrt{5}\right)$  

Which graph represents  $\frac{\left(y-1\right)^2}{16}-\frac{\left(x+1\right)^2}{4}=1$  

• 
• 
• 
• 

Tags

CCSS.HSA.CED.A.2

CCSS.HSA.REI.D.10

CCSS.HSF.IF.C.7

CCSS.HSG.GPE.A.3

Convert

 $16x^2-9y^2-64x+18y-89=0\$  to standard form

•  $\frac{\left(x-2\right)^2}{9}-\frac{\left(y-1\right)^2}{16}=1$  

•  $\frac{\left(x-2\right)^2}{9}-\frac{\left(y+1\right)^2}{16}=1$  

•  $\frac{\left(y-1\right)^2}{16}-\frac{\left(x-2\right)^2}{9}=1$  

•  $\frac{\left(y+1\right)^2}{16}-\frac{\left(x-2\right)^2}{9}=1$  

Tags

CCSS.HSG.GPE.A.3

Which graph represents  $16x^2-9y^2-64x+18y-89=0$  ?

• 
• 
• 
• 

Tags

CCSS.HSA.CED.A.2

CCSS.HSA.REI.D.10

What is the standard form equation of the hyperbola that has vertices at  $\left(1,\ -1\right)\ and\ \left(1,\ -5\right)$  and foci at  $\left(1,\ -3\pm2\sqrt{37}\right)$  ?

•  $\frac{\left(x-1\right)^2}{144}-\frac{\left(y+3\right)^2}{4}=1$  

•  $\frac{\left(y-1\right)^2}{4}-\frac{\left(x-3\right)^2}{144}=1$  

•  $\frac{\left(y+3\right)^2}{144}-\frac{\left(x-1\right)^2}{4}=1$  

•  $\frac{\left(y+3\right)^2}{4}-\frac{\left(x-1\right)^2}{144}=1$  

Write the standard form equation of the hyperbola that has vertices at (-2, -3) and (-8, -3) and an asymptote of  $y+3=\frac{8}{3}\left(x+5\right)$  .

•  $\frac{\left(x+5\right)^2}{64}-\frac{\left(y+3\right)^2}{9}=1$  

•  $\frac{\left(y+3\right)^2}{9}-\frac{\left(x+5\right)^2}{64}=1$  

•  $\frac{\left(y+3\right)^2}{64}-\frac{\left(x+5\right)^2}{9}=1$  

•  $\frac{\left(x+5\right)^2}{9}-\frac{\left(y+3\right)^2}{64}=1$  

 Write the standard form equation of the hyperbola that has foci at  $\left(-10,\ -8\pm\sqrt{157}\right)$  and asymptotes at  $y+8=\pm\frac{11}{6}\left(x+10\right)$  .

•  $\frac{\left(x+10\right)^2}{121}-\frac{\left(y+8\right)^2}{36}=1$  

•  $\frac{\left(y+8\right)^2}{121}-\frac{\left(x+10\right)^2}{36}=1$  

•  $\frac{\left(y+8\right)^2}{36}-\frac{\left(x+10\right)^2}{121}=1$  

•  $\frac{\left(x+10\right)^2}{36}-\frac{\left(y+8\right)^2}{121}=1$