How do you write an equation for a hyperbola with vertices (1, 3) and (-5, 3), and foci (3, 3) and (-7, 3)?

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Answer 1 · 2016-10-30T03:23:02

The equation is:

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

Please notice that the vertices are of the forms:

$(h - a, k)$ and $(h + a,k)$ specifically $(-5,3)$ and $(1, 3)$

The same information can be deduced from the foci, which have the forms:

$(h - c, k)$ and $(h + c, k)$ specifically $(-7, 3)$ and $(3, 3)$

The standard form for the equation of a hyperbola, where the vertices and foci have these properties, is the horizontal transverse axis form:

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

$k = 3$ by observation:

$(x - h)^2/a^2 - (y - 3)^2/b^2 = 1$

Compute h and a:

$-5 = h - a$ and $1 = h + a$

$2h = -4$

$h = -2$

$a = 3$

$(x - -2)^2/3^2 - (y - 3)^2/b^2 = 1$

To complete the equation, we only need the value of b but, to find the value of b, we must, first, find the value of c:

Using the $(h + c, k)$ form for the focus point, $(3,3)$ , we substitute -2 for h, set the right side equal to 3, and then solve for c:

$-2 + c = 3$

$c = 5$

Solve for b, using the equation $c^2 = a^2 + b^2$ :

$5^2 = 3^2 + b^2$

$b = 4$

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