# How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola y^2=36+4x^2?

Jul 20, 2017

#### Explanation:

The standard form of a hyperbola is either

${x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 1$

or

$\textcolor{b l u e}{{y}^{2} / {a}^{2} - {x}^{2} / {b}^{2} = 1}$

Our hyperbola is of the second form and has a vertical transverse axis, which means the foci and vertices are on the y-axis.

$\textcolor{red}{\text{Short cuts}} :$

Vertices will be at $\left(0 , a\right)$ and $\left(0 , - a\right)$

Asymptotes will be $y = \pm \left(\frac{a}{b}\right) x$

Foci are at $\left(0 , \pm c\right)$, where $c = \sqrt{{a}^{2} + {b}^{2}}$

So you can plug $a$ and $b$ into the above formulae, but I will give a bit of an explanation below.

First, rewrite the equation in standard form so you can find $a$ and $b$:

${y}^{2} = 36 + 4 {x}^{2} \Rightarrow {y}^{2} - 4 {x}^{2} = 36 \Rightarrow {y}^{2} / {6}^{2} - {x}^{2} / {3}^{2} = 1$

$\therefore a = 6$ and $b = 3$

1) The vertices are the turning points. These are found by setting x and y equal to zero.

y-intercepts (set x = 0):

${y}^{2} = 36 + 4 {\left(0\right)}^{2} \Rightarrow y = \pm 6$

There are no real x-intercepts as setting y to zero leads you to the square root of a negative number.

So we have vertices at:

$\left(0 , 6\right)$ and $\left(0 , - 6\right)$

2) For the asymptotes, solve the equation for y and look at the behaviour as x approaches $\pm \infty$. For large positive and negative values of x the hyperbola is essentially behaving like a straight line. I.e. it is asymptoting towards straight lines.

${y}^{2} = 36 + 4 {x}^{2} \Rightarrow y = \pm \sqrt{36 + 4 {x}^{2}}$

As x gets really large, we can ignore the 36, as it is essentially zero when compared with $\infty$. The equation then becomes

$y = \pm \sqrt{4 {x}^{2}} = \pm 2 x$

These are now the equations of the asymptotes:

$y = 2 x$

$y = - 2 x$

3) Foci equation:

${a}^{2} + {b}^{2} = {c}^{2}$

Solve for c to find the y-coordinates:

$c = \pm \sqrt{{a}^{2} + {b}^{2}} = \pm \sqrt{{6}^{2} + {3}^{2}} = \pm \sqrt{45} = \pm 3 \sqrt{5}$

Foci coordinates:

$\left(0 , 3 \sqrt{5}\right)$ and $\left(0 , - 3 \sqrt{5}\right)$

Now have a look at the graph, you can see that the foci and vertices are on the y-axis. You can also see that as x approaches $\pm \infty$ it asymptotes towards the two straight lines. 