Question

Prove that the hyperbola

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

has the two oblique asymptotes$y=\frac{a}{b}x$and$y=-\frac{a}{b}x$

Short answer

When $x\to \pm \infty$the term $\frac{b^2}{x^2}$become zero. Therefore, what remains is $y=\pm \frac{a}{b}x$which is exactly what you needed to prove.

Step-by-step solution

Step 1. Given Information

We prove that a hyperbola given with

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

has the line

$y=\pm \frac{a}{b}x$

for its asymptotes

Step 2. Concept

We want to express the variable $y$and then see what value does that expression becomes when we let$x\to \pm \infty$

Step 3. Proving the statement

The first step is to solve the given equation for $y$

$$\begin{gathered} \frac{y^2}{a^2}-\frac{x^2}{b^2}=1 \\ \frac{y^2}{a^2}=1+\frac{x^2}{b^2} \\ {y}^2=a^2\left[1+\frac{x^2}{b^2}\right] \\ y=\pm a\sqrt{1+\frac{x^2}{b^2}} \end{gathered}$$

Second step is to try to get the most similar expression as the wanted one. Therefore, from the previous equation you can get

$y=\pm a\sqrt{\frac{x^2}{b^2}\left(\frac{b^2}{x^2}+1\right)}$

and from this

$y=a\frac{x}{b}\left[\frac{b^2}{x^2}+1\right]$or $y=\frac{a}{b}x\left[\frac{b^2}{x^2}+1\right]$

which looks a lot like the wanted term. What is left to do is to observe the value of the term inside the parenthesis when$x\to \pm \infty$

Step 4. Hence Proved

When $x\to \pm \infty$the term

$\frac{b^2}{x^2}$

become zero.

Therefore, what remains is

$y=\pm \frac{a}{b}x$