Step 1

Given: center (-9, 1)

distance between \(\displaystyle{F}{1}{\quad\text{and}\quad}{F}{2}={20}\) units

distance between \(\displaystyle{C}{V}{1}{\quad\text{and}\quad}{C}{V}{2}={4}\) units

orientation = vertical

Step 2

Standard equation of hyperbola

\(\displaystyle\frac{{{\left({y}-{k}\right)}^{2}}}{{a}^{2}}-\frac{{{\left({x}-{h}\right)}^{2}}}{{b}^{2}}={1}\)

where (h,k) is centre and orientation is vertical

We know distance between \(\displaystyle{F}{1}{\quad\text{and}\quad}{F}{2}={2}{C}={20}\)

so \(\displaystyle{C}={10}\ \text{unit and}\ {c}^{2}={a}^{2}+{b}^{2}\)

here center is (-9, 1)

\(\displaystyle{h}=-{9}{\quad\text{and}\quad}{k}={1}\)

length of \(\displaystyle{C}{V}{1}\to{C}{V}{2}={2}{b}\)

\(\displaystyle{2}{b}={4}\) unit

\(\displaystyle{b}={2}\) unit

Now from \(\displaystyle{c}^{2}={a}^{2}+{b}^{2}\)

\(\displaystyle{\left({10}\right)}^{2}={a}^{2}+{\left({2}\right)}^{2}\)

\(\displaystyle{100}-{4}={a}^{2}\)

\(\displaystyle{a}=\sqrt{{96}}\)

Step 3

So the equation of hyperbola is

\(\displaystyle\frac{{\left({y}-{1}\right)}^{2}}{{96}}-\frac{{\left({x}+{9}\right)}^{2}}{{4}}={1}\)

where (-9, 1) is centre

Given: center (-9, 1)

distance between \(\displaystyle{F}{1}{\quad\text{and}\quad}{F}{2}={20}\) units

distance between \(\displaystyle{C}{V}{1}{\quad\text{and}\quad}{C}{V}{2}={4}\) units

orientation = vertical

Step 2

Standard equation of hyperbola

\(\displaystyle\frac{{{\left({y}-{k}\right)}^{2}}}{{a}^{2}}-\frac{{{\left({x}-{h}\right)}^{2}}}{{b}^{2}}={1}\)

where (h,k) is centre and orientation is vertical

We know distance between \(\displaystyle{F}{1}{\quad\text{and}\quad}{F}{2}={2}{C}={20}\)

so \(\displaystyle{C}={10}\ \text{unit and}\ {c}^{2}={a}^{2}+{b}^{2}\)

here center is (-9, 1)

\(\displaystyle{h}=-{9}{\quad\text{and}\quad}{k}={1}\)

length of \(\displaystyle{C}{V}{1}\to{C}{V}{2}={2}{b}\)

\(\displaystyle{2}{b}={4}\) unit

\(\displaystyle{b}={2}\) unit

Now from \(\displaystyle{c}^{2}={a}^{2}+{b}^{2}\)

\(\displaystyle{\left({10}\right)}^{2}={a}^{2}+{\left({2}\right)}^{2}\)

\(\displaystyle{100}-{4}={a}^{2}\)

\(\displaystyle{a}=\sqrt{{96}}\)

Step 3

So the equation of hyperbola is

\(\displaystyle\frac{{\left({y}-{1}\right)}^{2}}{{96}}-\frac{{\left({x}+{9}\right)}^{2}}{{4}}={1}\)

where (-9, 1) is centre