# How do you find the vertices, asymptote, foci and graph y^2/25-x^2/16=16?

Feb 22, 2017

Vertices: $\left(0 , \pm 20\right)$
Asymptotes: $y = \pm \frac{5}{4} x$
Foci: $\left(0 , \pm 4 \sqrt{41}\right)$

#### Explanation:

First put the hyperbola in standard form when the largest denominator is under the ${y}^{2}$: ${\left(y - k\right)}^{2} / {a}^{2} - {\left(x - h\right)}^{2} / {b}^{2} = 1$

Multiple all terms by $\frac{1}{16}$ to get a $1$ on the right-side:
${y}^{2} / \left(25 \cdot 16\right) - {x}^{2} / \left(16 \cdot 16\right) = 1$; ${y}^{2} / \left(400\right) - {x}^{2} / {16}^{2} = 1$; ${y}^{2} / {20}^{2} - {x}^{2} / {16}^{2} = 1$

Center: $\left(h , k\right) = \left(0 , 0\right)$

$a = 20 , b = 16$

${c}^{2} = {a}^{2} + {b}^{2} = 656 , c = \sqrt{656} = \sqrt{16 \cdot 41} = 4 \sqrt{41}$

Vertices: $\left(h , k \pm a\right) = \left(0 , \pm 20\right)$

Asymptotes: y-k = +-a/b(x-h); y = +-20/16x; y = +-5/4x

Foci: $\left(h , k \pm c\right) = \left(0 , \pm 4 \sqrt{41}\right)$

To graph:

1. Create a box centered around the center that is length $a$ vertically above and below the center and length $b$ horizontally above and below the center (use dashed lines).
2. Sketch the asymptote lines (use a dashed line).
3. Place dots at the vertices and foci.
4. Sketch a curve from each vertex, staying inside of the asymptote lines.
Graphing Hyperbolas