What is the asymptotes of+(x^2}{16}-\frac{y^2)/9 =1 ?

The asymptotes of+(x^2}{16}-\frac{y^2)/9 =1 is y=(3x)/4 ,\quad y=-(3x)/4

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asymptotes of+(x^2}{16}-\frac{y^2)/9 =1

asymptotes

+ \frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1

y = \frac{3 x}{4}, y = - \frac{3 x}{4}

Solution steps

y = \pm \frac{b}{a} (x - h) + k

Calculate Hyperbola properties

+ \frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 4, b = 3

y = \frac{3}{4} (x - 0) + 0, y = - \frac{3}{4} (x - 0) + 0

Refine

y = \frac{3 x}{4}, y = - \frac{3 x}{4}

9 x^{2} - 36 x - 4 y^{2} - 24 y - 36 = 0

\frac{y ^{2}}{64} - \frac{x ^{2}}{81} = 1

\frac{( y - 2 ) ^{2}}{9} - (x + 2)^{2} = 1

2 x^{2} - 3 y^{2} = 6

6 x^{2} - 36 y^{2} = 2304

What is the asymptotes of+(x^2}{16}-\frac{y^2)/9 =1 ?
The asymptotes of+(x^2}{16}-\frac{y^2)/9 =1 is y=(3x)/4 ,\quad y=-(3x)/4