How do you write an equation for the hyperbola with Foci: #(0,0)# and #(0,4)# Asmyptotes: #y= +-1/2x + 2#?

1 Answer
Jun 3, 2018

Answer:

Equation of vertical hyperbola is #(5(y-2)^2)/4-(5 x^2)/16=1#

Explanation:

Focii of hyperbola is at #(0,0) and (0,4)#

Center of the hyperbola is at #(0, 2) :. (h=0,k=2)#

So transverse axis is vertical , and equation of asymptotes

of vertical hyperbola are # y-k =+- a/b(x-h) #

Asymptotes: # y= +- 1/2 x+2; m=1/2 :. a/b= 1/2 :. b=2 a#

We know #c^2= a^2+b^2 ; c=4-2=2#

#:. 2^2= a^2+b^2 or 4 = a^2+(2 a)^2 or 5 a^2 =4# or

#a^2= 4/5 or a = 2/sqrt 5 :. b= 2a= 4 /sqrt 5#

The standard form of equation of vertical hyperbola is

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

#(y-2)^2/(4/5)-(x-0)^2/(16/5)=1# or

#(5(y-2)^2)/4-(5 x^2)/16=1#

graph{5(y-2)^2/4-5 x^2/16=1 [-12.66, 12.65, -6.33, 6.33]}