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Use the description below to find the equation of the hyperbola in standard form. Try making a sketch of the graph to help you. If a value is not an integer then type it as a decimal rounded to the nearest hundredth.Standard form:\frac{(y-k)^2}{a^2} -\frac{(x-h)^2}{b^2}=1 Description:Vertices at (0,6) and (0,-6) and one focus at (0,-8).The value for h is AnswerThe value for k is AnswerThe value for a is AnswerThe value for b is Answer

Use the description below to find the equation of the hyperbola in standard form. Try-example-1
User Jouhar
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1 Answer

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(y^(2))/(36)-(x^(2))/(28)=1

And the value of h=0

And the value of k is 0

And the value of a:2√7

And the value of b is 6

1) Since we were told that the vertices are (0,6) and (0,-6) and one focus is at (0,-8) we can write out the following equation to find the hyperbola:


\begin{gathered} h=0 \\ (k+8)^2=a^2+b^2 \\ \left(k-6\right)^2=b^2 \\ \left(k+6\right)^2=b^2 \\ (k+8)^2=a^2+\left(k-6\right)^2,h=0,k=0 \\ 64=a^2+36\Rightarrow64-36=a^2\Rightarrow a^2=28,a=√(28),a=2√(7) \\ \left(k-6\right)^2=b^2\Rightarrow b^2=36\operatorname{\Rightarrow}b=6 \end{gathered}

Note that to find a, and b we set h=0, k=0

2) Now, we can set the standard equation for this hyperbola:


\begin{gathered} (y^2)/(a^2)-(x^2)/(b^2)=1 \\ \\ (y^2)/(6^2)-(x^2)/((2√(7))^2)=1 \\ \\ (y^(2))/(6^(2))-(x^2)/((28))=1 \\ (y^2)/(36)-(x^2)/(28)=1 \\ \end{gathered}

User Somrlik
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