Question

What are the equations for the asymptotes of this hyperbola? y^2/36-x^2/121=1

Answer 1

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Answer to Question #89491 – Math – Calculus

Question

What are the equations for the asymptotes of this hyperbola? $y^2/36 - x^2/121 = 1$

Solution

$$\frac{y^2}{36} - \frac{x^2}{121} = 1$$

$\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$ is the standard equation

with center $(h,k)$, semi-axis $a$ and semi-conjugate-axis $b$.

$$\frac{(y - 0)^2}{6^2} - \frac{(x - 0)^2}{11^2} = 1$$

We get,

$$(h, k) = (0, 0), a = 6, b = 11$$

For hyperbola the asymptotes are $y = \pm \frac{a}{b} (x - h) + k$

$$\therefore y = \frac{6}{11} (x - 0) + 0, \quad y = -\frac{6}{11} (x - 0) + 0$$ $$y = \frac{6x}{11}, \quad y = -\frac{6x}{11}.$$