Question

question 1 of 10
what is the equation of the hyperbola with vertices at (0,-4) and (0,4) and foci at (0,-6) and (0,6)?
a. $\frac{x^{2}}{16}-\frac{y^{2}}{20}=1$
b. $\frac{y^{2}}{16}-\frac{x^{2}}{20}=1$
c. $\frac{x^{2}}{16}-\frac{y^{2}}{52}=1$
d. $\frac{x^{2}}{16}-\frac{y^{2}}{36}=1$

Explanation:

Step1: Determine the orientation and values of \(a\)

Since vertices \((0, - 4)\) and \((0,4)\) lie on the \(y -\)axis, the hyperbola has a vertical transverse - axis. The standard form of a vertical hyperbola is \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}} = 1\), and the distance from the center to a vertex is \(a\). Here, the center of the hyperbola is at the origin \((0,0)\) (mid - point of the vertices), and \(a = 4\), so \(a^{2}=16\).

Step2: Determine the value of \(c\) and then \(b^{2}\)

The foci are at \((0,-6)\) and \((0,6)\), and the distance from the center to a focus is \(c\). So \(c = 6\). We know the relationship \(c^{2}=a^{2}+b^{2}\) for a hyperbola. Substituting \(a = 4\) and \(c = 6\) into the equation \(c^{2}=a^{2}+b^{2}\), we get \(36=16 + b^{2}\). Solving for \(b^{2}\), we have \(b^{2}=c^{2}-a^{2}=36 - 16=20\).

Step3: Write the equation of the hyperbola

Substituting \(a^{2}=16\) and \(b^{2}=20\) into the standard form \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}} = 1\), we get \(\frac{y^{2}}{16}-\frac{x^{2}}{20}=1\).

Answer:

B. \(\frac{y^{2}}{16}-\frac{x^{2}}{20}=1\)