Question

Write an equation in standard form of the hyperbola described.

Vertex (10, 0); focus (-26, 0); center (0,0)

Answer 1

Check the picture below, so the hyperbola looks more or less like the one below, let's find the conjugate axis or namely the "b" component.

`\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad √( a ^2 + b ^2) \end{cases} \\\\[-0.35em] ~\dotfill`

`\begin{cases} h=0\\ k=0\\ a=10\\ c=26 \end{cases}\implies \cfrac{(x-0)^2}{10^2}-\cfrac{(y-0)^2}{b^2} \\\\\\ c^2=a^2+b^2\implies 26^2=10^2+b^2\implies 676=100+b^2\implies \underline{576=b^2} \\\\\\ \cfrac{(x-0)^2}{10^2}-\cfrac{(y-0)^2}{576}\implies \boxed{\cfrac{x^2}{100}-\cfrac{y^2}{576}}`