Thread: Help with hyperbola's equations.

1. Help with hyperbola's equations.

I have two question on an assignment of mine that I just can't seem to figure out, can you help me?

Write an equation of a hyperbola with vertices (1, 4) and (-5, 4), and foci (3, 4) and (-7, 4).

and

Write an equation of a hyperbola with vertices (8, 2) and (-4, 2), and foci (12,2) and (-8, 2).

2. so you know that the vertex are (1, 4),(-5, 4) so you find the center(x,y). with midpoint formula.

$\displaystyle (x,y) = \frac{x_1+x_2}{2} , \frac{y_1+y_2}{2}$

in your case center is (-2,4) so lets call this distance from center to vertex 'A'.
-5-(-2)=|-3| , 1-(-2)=|3|, A=3

now you can use hyperbola formula to derive standard form.
you know that A=3 and you have foci F at (3, 4),(-7, 4).

$\displaystyle | \sqrt{(X_1-FX_1)^2+(Y_1-FY_1)^2} - \sqrt{(X_2-FX_2)^2+(Y_2-FY_2)^2} | = 2A$

3. Hello, maramfaith24!

Write an equation of a hyperbola with vertices (1, 4) and (-5, 4)
and foci (3, 4) and (-7, 4).

You're expected to know the general forms of the hyperbola.

. . $\displaystyle \displaystyle \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} \:=\:1 \qquad \text{"horizontal" hyperbola: }\supset \subset$

. . $\displaystyle \displaystyle \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} \;=\;1 \qquad\text{"vertical" hyperbola: }\begin{array}{c}\cup \\ [-2mm]\cap \end{array}$

. . where $\displaystyle (h,k)$ is the center of the hyperbola
. . and the focal distance $\displaystyle \,c$ is given by: .$\displaystyle a^2+b^2 \:=\:c^2$ .[1]

The vertices are $\displaystyle (1,4)$ and $\displaystyle (\text{-}5,4)$, on a horizontal line.
. . We have a "horizontal" hyperbola.

The center is the midpoint of the vertices: .$\displaystyle (h,k) \,=\,(\text{-}2,4)$

The distance from the center $\displaystyle (\text{-}2,4)$ to a vertex $\displaystyle (1,4)$ is: .$\displaystyle 1 - (\text{-}2)$
. . Hence: $\displaystyle a = 3$

The distance from the center $\displaystyle (\text{-}2,4)$ to a focus $\displaystyle (3,4)$ is: .$\displaystyle 3- (\text{-}2)$
. . Hence: $\displaystyle c = 5$

Substitute into [1]: .$\displaystyle 3^2 + b^2 \:=\:5^2 \quad\Rightarrow\quad b^2 \,=\,16$

We know the center of the hyperbola: .$\displaystyle (h,k) \:=\:(\text{-}2,4)$
. . We know that: .$\displaystyle a^2 = 9,\;b^2 = 16$
. . And we know that the hyperbola is "horizontal".

$\displaystyle \displaystyle \text{The equation is: }\;\;\frac{(x+2)^2}{9} - \frac{(y-4)^2}{16} \;=\;1$