# 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

Determine the equation of a conic section...(Hyperbola) Given: center (-9, 1) distance between $F1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}F2=20$ units
distance between $CV1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}CV2=4$ units orientation = vertical
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dieseisB
Step 1
Given: center (-9, 1)
distance between $F1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}F2=20$ units
distance between $CV1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}CV2=4$ units
orientation = vertical
Step 2
Standard equation of hyperbola
$\frac{{\left(y-k\right)}^{2}}{{a}^{2}}-\frac{{\left(x-h\right)}^{2}}{{b}^{2}}=1$
where (h,k) is centre and orientation is vertical
We know distance between $F1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}F2=2C=20$
so
here center is (-9, 1)
$h=-9\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}k=1$
length of $CV1\to CV2=2b$
$2b=4$ unit
$b=2$ unit
Now from ${c}^{2}={a}^{2}+{b}^{2}$
${\left(10\right)}^{2}={a}^{2}+{\left(2\right)}^{2}$
$100-4={a}^{2}$
$a=\sqrt{96}$
Step 3
So the equation of hyperbola is
$\frac{{\left(y-1\right)}^{2}}{96}-\frac{{\left(x+9\right)}^{2}}{4}=1$
where (-9, 1) is centre
Jeffrey Jordon