Finding the Equation of a Hyperbola Centered at (h, k) Given its Foci and Vertices · Physics Playground

Example 3 from Algebra and Trigonometry, 12.2 The Hyperbola

Finding the Equation of a Hyperbola Centered at (h, k) Given its Foci and Vertices

What is the standard form equation of the hyperbola that has vertices at $(0, −2)$ and $(6, −2)$ and foci at $(−2, −2)$ and $(8, −2) ?$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
The y-coordinates of the vertices and foci are the same, so the transverse axis is parallel to the x-axis. Thus, the equation of the hyperbola will have the form

$\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1$

First, we identify the center, $(h, k) .$ The center is halfway between the vertices $(0, −2)$ and $(6, −2) .$ Applying the midpoint formula, we have

$(h, k) = (\frac{0 + 6}{2}, \frac{−2 + (−2)}{2}) = (3, −2)$
Next, we find $a^{2} .$ The length of the transverse axis, $2 a,$ is bounded by the vertices. So, we can find $a^{2}$ by finding the distance between the x-coordinates of the vertices.

$\begin{array}{l} 2 a = | 0 - 6 | \\ 2 a = 6 \\ a = 3 \\ a^{2} = 9 \end{array}$

Now we need to find $c^{2} .$ The coordinates of the foci are $(h \pm c, k) .$ So $(h - c, k) = (- 2, −2)$ and $(h + c, k) = (8, −2) .$ We can use the x-coordinate from either of these points to solve for $c .$ Using the point $(8, −2),$ and substituting $h = 3,$

$\begin{array}{l} h + c = 8 \\ 3 + c = 8 \\ c = 5 \\ c^{2} = 25 \end{array}$

Next, solve for $b^{2}$ using the equation $b^{2} = c^{2} - a^{2} :$

$\begin{array}{l} b^{2} = c^{2} - a^{2} \\ = 25 - 9 \\ = 16 \end{array}$

Finally, substitute the values found for $h, k, a^{2},$ and $b^{2}$ into the standard form of the equation.

$\frac{{(x - 3)}^{2}}{9} - \frac{{(y + 2)}^{2}}{16} = 1$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

What is the standard form equation of the hyperbola that has vertices $(1, −2)$ and $(1, 8)$ and foci $(1, −10)$ and $(1, 16) ?$

How did it go?