How do you write the equation for a hyperbola given vertices (9,-3) and (-5,-3), foci (2+-sqrt53,-3)(2±53,3)?

1 Answer
Feb 21, 2018

The equation of hyperbola is (x-2)^2/49-(y+3)^2/4=1(x2)249(y+3)24=1

Explanation:

Vertices are (9,-3) and (-5,-3)(9,3)and(5,3)

Foci are (2+sqrt53,-3) and (2-sqrt53,-3)(2+53,3)and(253,3)

By the Midpoint Formula, the center of the hyperbola occurs at the

point (2,-3); h=2, k=-3 :. a= 9-2=7; a^2=49 ;

c= 2+sqrt53 - 2= sqrt53:. c^2=53

b^2= c^2-a^2=53-49=4 :. b=2 . So, the hyperbola has a

horizontal transverse axis and the standard form of the

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

(x-2)^2/7^2-(y+3)^2/2^2=1 or

(x-2)^2/49-(y+3)^2/4=1

The equation of hyperbola is (x-2)^2/49-(y+3)^2/4=1 [Ans]