Question

1. determine the equation of the hyperbola with vertices at (±4,0) and foci at (±6,0).

\frac{x^{2}}{13}-\frac{y^{2}}{36}=1
\frac{y^{2}}{16}-\frac{x^{2}}{20}=1
\frac{x^{2}}{16}-\frac{y^{2}}{20}=1
\frac{y^{2}}{36}-\frac{x^{2}}{16}=1

1. determine the equation of the hyperbola with vertices at (0,±4) and foci at (0,±6).

\frac{x^{2}}{13}-\frac{y^{2}}{30}=1
\frac{y^{2}}{16}-\frac{x^{2}}{20}=1
\frac{x^{2}}{16}-\frac{y^{2}}{20}=1
\frac{y^{2}}{36}-\frac{x^{2}}{16}=1

Explanation:

Step1: Recall the standard - form of hyperbola equations

The standard form of a hyperbola with a horizontal transverse axis is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1$, and with a vertical transverse axis is $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}} = 1$. The distance from the center to a vertex is $a$, and the distance from the center to a focus is $c$. Also, $c^{2}=a^{2}+b^{2}$.

Step2: Solve for the first hyperbola (vertices $(\pm4,0)$ and foci $(\pm6,0)$)

Since the vertices are $(\pm4,0)$, the center of the hyperbola is at the origin $(0,0)$ and $a = 4$. Since the foci are $(\pm6,0)$, $c = 6$. Using the formula $c^{2}=a^{2}+b^{2}$, we substitute $a = 4$ and $c = 6$:

$$b^{2}=c^{2}-a^{2}=6^{2}-4^{2}=36 - 16=20$$

The equation of the hyperbola with a horizontal transverse axis is $\frac{x^{2}}{16}-\frac{y^{2}}{20}=1$.

Step3: Solve for the second hyperbola (vertices $(0,\pm4)$ and foci $(0,\pm6)$)

Since the vertices are $(0,\pm4)$, the center of the hyperbola is at the origin $(0,0)$ and $a = 4$. Since the foci are $(0,\pm6)$, $c = 6$. Using the formula $c^{2}=a^{2}+b^{2}$, we substitute $a = 4$ and $c = 6$:

$$b^{2}=c^{2}-a^{2}=6^{2}-4^{2}=36 - 16 = 20$$

The equation of the hyperbola with a vertical transverse axis is $\frac{y^{2}}{16}-\frac{x^{2}}{20}=1$.

Answer:

1. $\frac{x^{2}}{16}-\frac{y^{2}}{20}=1$
2. $\frac{y^{2}}{16}-\frac{x^{2}}{20}=1$