Question

Determine the equation of a conic section...(Hyperbola) Given: center (-9, 1) distance between displaystyle{F}{1}{quadtext{and}quad}{F}{2}={20} units distance between displaystyle{C}{V}{1}{quadtext{and}quad}{C}{V}{2}={4} units orientation = vertical

Conic sections
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asked 2021-02-05
Determine the equation of a conic section...(Hyperbola) 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

Answers (1)

2021-02-06
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
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