Question

What is the equation of the hyperbola with vertices at (-4,0) and (4,0) and foci at (-6,0) and (6,0)? .2 O 31 O A. 16 36 2 .ܐ 1 O B. 16 20 = v? 52 II 1 C. 16 2 ,2 y OD - 1 O D. 16 20

Answer 1

We are to find the equation of the hyperbola with vertices at (-4,0) and (4,0) and foci at (-6,0) and (6,0).

The equation is of the form;

`(x^2)/(a^2)\text{ - }(y^2)/(b^2)\text{ = 1}`
`\begin{gathered} \text{Vertex (a) = 4} \\ \text{Focal length (c) = 6} \end{gathered}`
`\begin{gathered} c^2=a^2+b^2 \\ 6^2=4^2+b^2 \\ 36=16+b^2 \\ b^2\text{ = 36 - 16} \\ b^2\text{ = 20} \\ b\text{ = }\sqrt[]{20} \end{gathered}`

`\begin{gathered} (x^2)/(a^2)\text{ - }(y^2)/(b^2)\text{ = 1} \\ \\ (x^2)/(4^2)\text{ - }\frac{y^2}{(\sqrt[]{20}^{})^2}\text{ = 1} \\ \\ \frac{x^2}{16^{}}\text{ - }\frac{y^2}{20^{}}\text{ = 1} \end{gathered}`