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Write the equation in standard form for the hyperbola with vertices (-2,0) and (2,0) and a conjugate axis of length 14

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Solution

- The equation of a hyperbola is given s:


\begin{gathered} ((x-h)^2)/(a)-((y-k)^2)/(b)=1 \\ \\ where, \\ coordinates\text{ of the vertices}=(h\pm a,k) \\ Length\text{ of conjugate axis}=2b \end{gathered}

- Thus, we can find that:


\begin{gathered} (\pm2,0)=(h\pm a,k) \\ \\ k=0 \\ \therefore h+a=2 \\ h-a=-2 \\ \text{ Subtract both equations, we have:} \\ 2a=4 \\ a=(4)/(2)=2 \\ \\ h+a=2 \\ h+2=2 \\ h=2-2=0 \\ \\ \text{ Thus, we have that the center of the hyperbola is: }(h,k)=(0,0) \\ \\ 2b=14 \\ \text{ Divide both sides by 2} \\ b=(14)/(2)=7 \end{gathered}

Final Answer

The equation of the parabola is:


(x^2)/(2^2)-(y^2)/(7^2)=1

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