91Ó°ÊÓ

Hyperbola: center $\left(2,2\right)$, vertex $\left(2,4\right)$, contains the point$\left(2+\sqrt{10},5\right)$

Given that center $\left(2,2\right)$therefore $h=2andk=2$,

We will use vertical hyperbola form $\frac{(y-k{)}^2}{a^2}-\frac{(x-h{)}^2}{b^2}=1$

The vertex of a hyperbola is of the form $\left(h,kÂ\pm a\right)$

Using the vertex $\left(2,4\right)$to find the value of $a$,

$$\begin{gathered} \left(h,kÂ\pm a\right)=\left(2,4\right) \\ \left(h,kÂ\pm a\right)=\left(2,2+2\right) \\ a=2 \\ {a}^2=4 \end{gathered}$$

Substitute the value of $a,h,k$and the point role="math" localid="1648442989592" $\left(x,y\right)=\left(2+\sqrt{10},5\right)$in the equation of the vertical hyperbola, role="math" localid="1648443343813" $\frac{(y-k{)}^2}{a^2}-\frac{(x-h{)}^2}{b^2}=1$and simplify.

role="math" localid="1648443249575" $$\begin{gathered} \frac{(y-k{)}^2}{a^2}-\frac{(x-h{)}^2}{b^2}=1 \\ \frac{(5-2{)}^2}{2^2}-\frac{(2+\sqrt{10}-2{)}^2}{b^2}=1 \\ \frac{(3{)}^2}{2^2}-\frac{(\sqrt{10}{)}^2}{b^2}=1 \\ \frac{9}{4}-\frac{10}{b^2}=1 \\ -\frac{10}{b^2}=-\frac{5}{4} \\ {b}^2=8 \\ b=2\sqrt{2} \end{gathered}$$

Substitute the value of $b^2=8,a^2=4,h=2\text{and}k=2$in the equation of vertical hyperbola

$\frac{(y-k{)}^2}{a^2}-\frac{(x-h{)}^2}{b^2}=1$we get,

role="math" localid="1648443391484" $$\begin{gathered} \frac{(y-k{)}^2}{a^2}-\frac{(x-h{)}^2}{b^2}=1 \\ \frac{(y-2{)}^2}{4}-\frac{(x-2{)}^2}{8}=1 \end{gathered}$$

The graph of the hyperbola $\frac{(y-2{)}^2}{4}-\frac{(x-2{)}^2}{8}=1$is