Question

question 2
1 pts
what are the asymptotes of the hyperbola given by the equation 1296y² - x² = 81?
o y = ±1/36x
o y = ±36x
o y = ±2/3x
o y = ±3/2x

Explanation:

Step1: Rewrite the hyperbola equation in standard form

First, rewrite $1296y^{2}-x^{2}=81$ as $\frac{y^{2}}{\frac{81}{1296}}-\frac{x^{2}}{81}=1$, which simplifies to $\frac{y^{2}}{\frac{1}{16}}-\frac{x^{2}}{81}=1$. Here, $a^{2}=\frac{1}{16}$ so $a = \frac{1}{4}$, and $b^{2}=81$ so $b = 9$.

Step2: Use the formula for asymptotes of a vertical - transverse - axis hyperbola

The formula for the asymptotes of a hyperbola of the form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$ is $y=\pm\frac{a}{b}x$. Substitute $a=\frac{1}{4}$ and $b = 9$ into the formula: $y=\pm\frac{\frac{1}{4}}{9}x=\pm\frac{1}{36}x$.

Answer:

$y=\pm\frac{1}{36}x$