1. ## Asymptotes for hyperbolas

Find the asymptotes for the hyperbola $
\frac{2x^2}{3} - \frac{y^2}{2}=1
$

I got the following:
$
y=\frac{\sqrt{6}}{3}
$
and $y=\frac{-\sqrt{6}}{3}$

However the answer is $y=\pm \frac{4}{\sqrt{3}}$

Where did I go wrong? Please, any help would be appreciated!

2. Originally Posted by Joker37
Find the asymptotes for the hyperbola $
\frac{2x^2}{3} - \frac{y^2}{2}=1
$

I got the following:
$
y=\frac{\sqrt{6}}{3}
$
and $y=\frac{-\sqrt{6}}{3}$

However the answer is $y=\pm \frac{4}{\sqrt{3}}$

Where did I go wrong? Please, any help would be appreciated!
$\displaystyle\frac{2x^2}{3}-\frac{y^2}{2}=1\Rightarrow\ y^2=\frac{4x^2}{3}-2\Rightarrow\ y=\pm\frac{\sqrt{4x^2-6}}{\sqrt{3}}$

$\Rightarrow\ y=\displaystyle\pm\frac{x\sqrt{4-\frac{6}{x^2}}}{\sqrt{3}}$

and as $x\rightarrow\infty$, this is $\displaystyle\ y=\pm\frac{2x}{\sqrt{3}}$

3. From Archie's post to know where the function is undefined solve for $\displaystyle 4x^2-6<0 \implies -\frac{\sqrt{3}}{\sqrt{2}}

4. Do you understand that it is impossible to say "where you went wrong" if you don't show what you did?

In any case, a hyperbola having the coordinate axes as axes of symmetry (any hyperbola of the form $\frac{x^2}{a^2}- \frac{y^2}{b^2}= 1$ or $\frac{y^2}{b^2}- \frac{x^2}{a^2}= 1$) has slant asymptotes, not vertical and horizontal asymptotes.

No, the anwer is not $y= \pm\frac{4}{\sqrt{3}}$. Neither $y= \pm\frac{\sqrt{6}}{3}$ nor $y= \pm\frac{4}{\sqrt{3}}$ is correct.

Here is one way to think about it. The graph of the hyperbola gets closer and closer to the asymptotes as x and y get larger and larger. For very, very large x and y, both $\frac{2x^2}{3}$ and $\frac{y^2}{2}$ are very, very large so that "1" can be ignored compared to them. That is, for very, very large x and y, $\frac{2x^2}{3}- \frac{y^2}{2}= 0$, approximately. The left side is a "difference of squares" and factors as $\left(\frac{x\sqrt{2}}{\sqrt{3}}+ \frac{y}{\sqrt{2}}\right)\left(\frac{x\sqrt{2}}{\s qrt{3}}- \frac{y}{\sqrt{2}}\right)= 0$.

So we must have $\frac{x\sqrt{2}}{\sqrt{3}}+ \frac{y}{\sqrt{2}}= 0$ and $\frac{x\sqrt{2}}{\sqrt{3}}+ \frac{y}{\sqrt{2}}= 0$, the equation of the two straight line asymptotes. Solving for y, we have $y= \pm\frac{2}{\sqrt{3}}x$ as Archie Meade says.

5. Originally Posted by Archie Meade
$\displaystyle\frac{2x^2}{3}-\frac{y^2}{2}=1\Rightarrow\ y^2=\frac{4x^2}{3}-2\Rightarrow\ y=\pm\frac{\sqrt{4x^2-6}}{\sqrt{3}}$

$\Rightarrow\ y=\displaystyle\pm\frac{x\sqrt{4-\frac{6}{x^2}}}{\sqrt{3}}$

and as $x\rightarrow\infty$, this is $\displaystyle\ y=\pm\frac{2x}{\sqrt{3}}$
Technically, the generalised binomial theorem should be used to find this limit.

6. Hello, Joker37!

$\text{Find the asymptotes for the hyperbola: }\:\dfrac{2x^2}{3} - \dfrac{y^2}{2}\:=\:1$

We're expected to be familiar with this fact:

Given the hyperbola: . $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} \:=\:1$, the asymptotes are: . $y \;=\;\pm\dfrac{b}{a}x$

The equation is: . $\displaystyle \frac{x^2}{\frac{3}{2}} - \frac{y}{2} \;=\;1 \quad\Rightarrow\quad \frac{x^2}{\left(\sqrt{\frac{3}{2}}\right)^2} - \frac{y^2}{(\sqrt{2})^2} \;=\;1$

Hence: . $a = \sqrt{\frac{3}{2}},\;b = \sqrt{2}$

Therefore: . $y \;=\;\pm\dfrac{\sqrt{2}}{\sqrt{\frac{3}{2}}}x \quad\Rightarrow\quad y \;=\;\pm\dfrac{2}{\sqrt{3}}x$