Question

Write equation of hyperbola from key features

Answer 1

Check the picture below, so the hyperbola looks more or less like so.

the center of the hyperbola is half-way between the foci, so namely at (-2 , -1) as you see there, since we know the covertices, half that distance is the "b" component, so all we're really missing is the "a" component, but we know what "c", since it's just the distance from a foci to the center, so

`\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=-2\\ k=-1\\ b=15 \end{cases}\implies \cfrac{( ~~ x- (-2) ~~ )^2}{ a^2}-\cfrac{( ~~ y- (-1) ~~ )^2}{ 15^2}=1`

`\stackrel{\textit{we know that}}{c=17}\qquad 17=√(a^2+15^2)\implies 17^2=a^2+15^2\implies 17^2-15^2=a^2 \\\\\\ √(17^2-15^2)=a\implies √(64)=a\implies 8=a \\\\[-0.35em] ~\dotfill\\\\ \cfrac{( ~~ x- (-2) ~~ )^2}{ 8^2}-\cfrac{( ~~ y- (-1) ~~ )^2}{ 15^2}=1\implies {\Large \begin{array}{llll} \cfrac{(x+2)^2}{64}-\cfrac{(y+1)^2}{225}=1 \end{array}}`