How do you write the equation of a hyperbola in standard form given Foci: (3,+-2) and Asymptotes: y = +-2(x-3)?

Question

How do you write the equation of a hyperbola in standard form given Foci: (3,+-2) and Asymptotes: y = +-2(x-3)?

1 Answer

Impact of this question

3742 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-31T07:05:47

The equation of the hyperbola is $\frac{{(y)}^{2}}{\frac{16}{5}} - \frac{{(x - 3)}^{2}}{\frac{4}{5}} = 1$ $(y)^2/(16/5)-(x-3)^2/(4/5)=1$

Explanation:

The standard equation of a hyperbola is
$\frac{{(y - k)}^{2}}{b^{2}} - \frac{{(x - h)}^{2}}{a^{2}} = 1$ $(y-k)^2/b^2-(x-h)^2/a^2=1$
The slopes of the asymptotes are $\pm \frac{b}{a}$ $+-b/a$
so $\pm \frac{b}{a} = \pm 2$ $+-b/a=+-2$ $\Rightarrow$ $=>$ $b = 2 a$ $b=2a$
The foci are $h, k \pm c$ $h,k+-c$
So $h = 3$ $h=3$
and $k + c = 2$ $k+c=2$ and $k - c = - 2$ $k-c=-2$
So from the eqautions $k = 0$ $k=0$ and $c = 2$ $c=2$
$c^{2} = a^{2} + b^{2}$ $c^2=a^2+b^2$
Therefore $4 = a^{2} + 4 a^{2} = 5 a^{2}$ $4=a^2+4a^2=5a^2$ $\Rightarrow$ $=>$ $a^{2} = \frac{4}{5}$ $a^2=4/5$
And $b^{2} = 4 a^{2} = \frac{16}{5}$ $b^2=4a^2=16/5$

So the equation of the hyperbola is
$\frac{{(y)}^{2}}{\frac{16}{5}} - \frac{{(x - 3)}^{2}}{\frac{4}{5}} = 1$ $(y)^2/(16/5)-(x-3)^2/(4/5)=1$

Science

Math

Humanities

... and beyond

How do you write the equation of a hyperbola in standard form given Foci: (3,+-2) and Asymptotes: y = +-2(x-3)?

1 Answer

Explanation:

Related questions

Impact of this question