Question

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

Answer 1

Answer:

D.

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

Step-by-step explanation:

we are given

we can use standard equation of hyperbola

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$

where

center=(h,k)

center at origin

so, h=0 and k=0

vertex is

$(0,\sqrt{12})$

we can use formula

vertices: (h, k + a)

we get

$k+a=\sqrt{12}$

we can plug k=0

$a=\sqrt{12}$

now, we can plug these values

$\frac{(y-0)^2}{(\sqrt{12})^2}-\frac{(x-0)^2}{b^2}=1$

now, we are given it passes through $(2\sqrt{3} ,6)$

so, we have

$x=2\sqrt{3},y=6$

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

$\frac{(6-0)^2}{(\sqrt{12})^2}-\frac{(2\sqrt{3}-0)^2}{b^2}=1$

and we get

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

we can solve for b

and we get

$b=\sqrt{6}$

now, we can plug these values

$\frac{(y-0)^2}{(\sqrt{12})^2}-\frac{(x-0)^2}{(\sqrt{6})^2}=1$

we can simplify it

and we get

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

Answer 2

Answer:

D.

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