Question

Which is the equation of a hyperbola centered at the origin with vertex (0,sqrt12) that passes through (2sqrt3,6)

A. y^2/36-x^2/12=1
B. y^2/12-x^2/36=1
C. y^2/6-x^2/12=1
D. y^2/12-x^2/6=1

Answer 1

D.

\frac{y^2}{12}-\frac{x^2}{6}=1

Answer 2

D.

`(y^2)/(12)-(x^2)/(6)=1`

Explanation:

we are given

we can use standard equation of hyperbola

`((y-k)^2)/(a^2)-((x-h)^2)/(b^2)=1`

where

center=(h,k)

center at origin

so, h=0 and k=0

vertex is

`(0,√(12))`

we can use formula

vertices: (h, k + a)

we get

`k+a=√(12)`

we can plug k=0

`a=√(12)`

now, we can plug these values

`((y-0)^2)/((√(12))^2)-((x-0)^2)/(b^2)=1`

now, we are given it passes through
`(2√(3) ,6)`

so, we have

`x=2√(3),y=6`

we can plug these values and then we can solve for b

`((6-0)^2)/((√(12))^2)-((2√(3)-0)^2)/(b^2)=1`

and we get

`36b^2-144=12b^2`

we can solve for b

and we get

`b=√(6)`

now, we can plug these values

`((y-0)^2)/((√(12))^2)-((x-0)^2)/((√(6))^2)=1`

we can simplify it

and we get

`(y^2)/(12)-(x^2)/(6)=1`