Emily-Jane Bray

2021-02-05

Determine the equation of a conic section...(Hyperbola) Given: center (-9, 1) distance between $F1{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}F2=20$ units

distance between $CV1{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}CV2=4$ units orientation = vertical

dieseisB

Skilled2021-02-06Added 85 answers

Step 1

Given: center (-9, 1)

distance between$F1{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}F2=20$ units

distance between$CV1{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}CV2=4$ units

orientation = vertical

Step 2

Standard equation of hyperbola

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

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

We know distance between$F1{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}F2=2C=20$

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

here center is (-9, 1)

$h=-9{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}k=1$

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

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

where (-9, 1) is centre

Given: center (-9, 1)

distance between

distance between

orientation = vertical

Step 2

Standard equation of hyperbola

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

We know distance between

so

here center is (-9, 1)

length of

Now from

Step 3

So the equation of hyperbola is

where (-9, 1) is centre

Jeffrey Jordon

Expert2021-10-27Added 2605 answers

Answer is given below (on video)

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