The equation for a hyperbola that opens up and down has the following general form:
[tex]\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1[/tex]Where the foci of the hyperbola are located at (h,k+c) and (h,k-c) with c given by:
[tex]c^2=a^2+b^2[/tex]And asymptotes with slopes given by a/b and -a/b.
The hyperbola with the equation that we have to find has these two foci:
[tex](3,2-\sqrt{26})\text{ and }(3,2+\sqrt{26})[/tex]This means that:
[tex]\begin{gathered} (h,k-c)=(3,2-\sqrt{26}) \\ (h,k+c)=(3,2+\sqrt{26}) \end{gathered}[/tex]So we get h=3, k=2 and c=√26.
The slope of the asymptotes have to be 5 and -5 which means that:
[tex]\frac{a}{b}=5[/tex]Using the value of c we have:
[tex]c^2=26=a^2+b^2[/tex]So we have two equation for a and b. We can take the first one and multiply b to both sides:
[tex]\begin{gathered} \frac{a}{b}\cdot b=5b \\ a=5b \end{gathered}[/tex]And we use this in the second equation:
[tex]\begin{gathered} 26=(5b)^2+b^2=25b^2+b^2 \\ 26=26b^2 \end{gathered}[/tex]We divide both sides by 26:
[tex]\begin{gathered} \frac{26}{26}=\frac{26b^2}{26} \\ b^2=1 \end{gathered}[/tex]Which implies that b=1. Then a is equal to:
[tex]a=5b=5\cdot1=5[/tex]AnswerNow that we have found a, b, h and k we can write the equation of the hyperbola. Then the answer is:
[tex]\frac{(y-2)^2}{5^2}-\frac{(x-3)^2}{1^2}=1[/tex]