Question

Find the standard form of the equation of the hyperbola satisfying the given conditions.

x-intercepts ±24; foci at (-25,0) and (25,0).

Answer 1

Check the picture below, so the hyperbola looks more or less like so, with a center at the origin and a "c" distance from the center to a focus point of 25 units.

`\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=24\\ c=25 \end{cases}\implies \cfrac{(x- 0)^2}{ 24^2}-\cfrac{(y-0)^2}{ b^2}=1`

`c=√(a^2+b^2)\implies c^2=a^2+b^2\implies c^2-a^2=b^2\implies 25^2 - 24^2 = b^2 \\\\\\ 49=b^2\hspace{7em}\cfrac{(x- 0)^2}{ 24^2}-\cfrac{(y-0)^2}{ b^2}=1\implies {\Large \begin{array}{llll} \cfrac{x^2}{576}-\cfrac{y^2}{49}=1 \end{array}}`