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

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Answer 1 · 2018-02-21T12:59:33

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

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

Foci are # (2+sqrt53,-3) and (2-sqrt53,-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]