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]

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