Hyperbola

It is the set of all points in the plane such that the difference of their distances from two fixed points (foci) is constant.

• circle

• ellipse

• hyperbola

• parabola

What is the length of the transverse axis of a hyperbola whose equation is $\frac{y^2}{25}-\frac{x^2}{16}=1$  . (Transverse axis is twice a).

• 8

• 10

• 16

• 25

What is the length of the conjugate axis of a hyperbola whose equation is $\frac{y^2}{25}-\frac{x^2}{16}=1$  . (Conjugate axis is twice b).

• 8

• 10

• 16

• 25

What is orientation of a hyperbola whose equation is $\frac{y^2}{25}-\frac{x^2}{16}=1$  . 

• diagonal

• horizontal

• vertical

• wavy

What is the coordinate of the center of a hyperbola whose equation is $\frac{\left(x-1\right)^2}{4}-\frac{\left(y-1\right)^2}{12}=1$  . 

• (1, 1)

• (-1, -1)

• (4, 12)

• (12, 4)

What are the coordinates of the endpoints of conjugate axis of a hyperbola whose equation is $\frac{\left(x-1\right)^2}{4}-\frac{\left(y-1\right)^2}{12}=1$  . 

• (1, 13) and (1, -11)

•  $\left(1,\ 1+2\sqrt{3}\right)\ \ and\ \ \ \left(1,\ 1-2\sqrt{3}\right)$  
  

• (5, 1) and (-3, 1)

• (3, 1) and (-1, 1)

What are the coordinates of the vertices of a hyperbola whose equation is $\frac{\left(x-1\right)^2}{4}-\frac{\left(y-1\right)^2}{12}=1$  . 

• (13, 1) and (-11, 1)

•  $\left(1+2\sqrt{3},\ 1\right)\ \ and\ \ \ \left(1-2\sqrt{3},\ 1\right)$  
  

• (5, 1) and (-3, 1)

• (3, 1) and (-1, 1)

What are the coordinates of the foci of a hyperbola whose equation is $\frac{\left(x-1\right)^2}{4}-\frac{\left(y-1\right)^2}{12}=1$  . 

• (13, 1) and (-11, 1)

•  $\left(1+2\sqrt{3},\ 1\right)\ \ and\ \ \ \left(1-2\sqrt{3},\ 1\right)$  
  

• (5, 1) and (-3, 1)

• (3, 1) and (-1, 1)

What is the equation of asymptotes of a hyperbola whose equation is $\frac{\left(x-1\right)^2}{4}-\frac{\left(y-1\right)^2}{12}=1$  . 

•  $y=1\pm\frac{2}{2\sqrt{3}}\left(x-1\right)$  
  

•  $y=1\pm\frac{2\sqrt{3}}{2}\left(x-1\right)$  
  

Determine the equation of the hyperbola.

• $\frac{\left(y-1\right)^2}{4}-\frac{\left(x-1\right)^2}{12}=1$

• $\frac{y^2}{9}-\frac{x^2}{16}=1$

• $\frac{x^2}{16}-\frac{y^2}{9}=1$

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