Given that the equation of asymptotes to the hyperbola be:

$y=\pm\frac{3x}{2}$ and $b=4$

How to find the equation of hyperbola?

I know that asymtotes have the equation $y=\pm\frac{bx}{a}$, comparing and solving we get $a=\frac{8}{3}$

But in the exercise there are two answers given :

$\frac{9x^2}{64}-\frac{y^2}{16}=1$ and $\frac{y^2}{36}-\frac{x^2}{16}=1$.

How are there tw0 answers. Please Help. Thanks.

  • $\begingroup$ I believe you mean the asymptotes to be $y= \pm \frac 32 x$: you left out the $x$. $\endgroup$ – Rory Daulton Sep 1 '14 at 14:35
  • 1
    $\begingroup$ It could be either $$\dfrac{b}{a} = \dfrac{3}{2}$$ or $$\dfrac{a}{b} = \dfrac{3}{2}$$ $\endgroup$ – rrr Sep 1 '14 at 14:37
  • $\begingroup$ @user172209 i dont understand ?? $\endgroup$ – Shobhit Sep 1 '14 at 14:38
  • $\begingroup$ @RoryDaulton thanks $\endgroup$ – Shobhit Sep 1 '14 at 14:38
  • $\begingroup$ How do you know whether the hyperbola is horizontal or vertical ? $\endgroup$ – rrr Sep 1 '14 at 14:39


If the given asymptotes are for horizontal hyperbola $$\dfrac{b}{a} = \dfrac{3}{2}$$

otherwise for vertical :

$$\dfrac{a}{b} = \dfrac{3}{2}$$


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