Find the equations for the asymptotes of the hyperbola y216−x29=1\frac{y^2}{16}-\frac{x^2}{9}=116y2−9x2=1.

y = ± 4 3 x y=\pm\frac43x y = ± 3 4 x

Find the equations for the asymptotes of the hyperbola y 2 16 − x 2 9 = 1 \frac{y^2}{16}-\frac{x^2}{9}=1 16 y 2 − 9 x 2 = 1 .

Welcome back everyone. So let's give this problem a try. In this problem, we're asked to find the equations for the asymptotes of this hyperbola. Now Now what I can do to solve this problem is recognize the type of hyperbola that we have. I see that a y squared shows up first, and that means that we are dealing with a vertical hyperbola. So the equation that we need for the asymptotes is y is equal to plus or minus a over b times x. Now if you did not remember this equation off the top of your head, don't worry about it because there is a way to do this. If you recognize that we have a vertical hyperbola, then the hyperbola is going to be oriented on the y axis. So it's going to look something like this, where you have the curves diverging away from each other. Now, the distance to either one of these curves from the center, this would be a distance of a, and then there would be another distance we would define which is b. And what you could do is connect these distances with a box, and this would allow you to figure out the asymptotes that you're looking for. So if we connect these with a box, what we could do is draw the asymptotes through the corners. So one asymptote would go right through here, and another asymptote would go right through there. Now, the reason why this helps us to determine this equation is because notice for this box that we can find the slope of this line, which is going to be rise over run. And notice how rise over run is just a over b. So our line would be y is equal to plus or minus the slope, which is a over b times x. So that is how we got the equation up here, in case you're wondering. But now that we have this equation, let's see if we can solve the problem and find the asymptotes. So to find the asymptotes, I need a and b. And a squared is going to be the first thing that we see in the denominator of this equation, and b squared is going to be the second number that we see. So since a squared is equal to 16, that means that a is going to be the square root of 16, which is 4. And since b squared is equal to 9, b is going to be the square root of 9, which is 3. So if a is equal to 4, and b is equal to 3, we can use this equation to get that y is equal to plus or minus a, which is 4, divided by b, which is 3 times x. And that is going to be the equation for each of these asymptotes, where the positive 4 thirds x is associated with this upward sloping line, and the negative 4 thirds x is associated with the downward sloping line. This is the equation we're looking for, and that's the answer to this problem. So hope you found this video helpful, thanks for watching.

Find the equations for the asymptotes of the hyperbola y 2 16 − x 2 9 = 1 \frac{y^2}{16}-\frac{x^2}{9}=1 16 y 2 − 9 x 2 = 1 .

y = ± 4 3 x y=\pm\frac43x y = ± 3 4 x

y = ± 9 16 x y=\pm\frac{9}{16}x y = ± 16 9 x

y = ± 16 9 x y=\pm\frac{16}{9}x y = ± 9 16 x

y = ± 3 4 x y=\pm\frac34x y = ± 4 3 x

19. Conic Sections

Hyperbolas at the Origin

Precalculus

19. Conic Sections

Hyperbolas at the Origin

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Multiple Choice

Find the equations for the asymptotes of the hyperbola $\frac{y^2}{16}-\frac{x^2}{9}=1$ .

$y=\pm\frac{9}{16}x$

$y=\pm\frac{16}{9}x$

$y=\pm\frac34x$

$y=\pm\frac43x$

Verified step by step guidance

Identify the standard form of the hyperbola equation. The given equation is $ \frac{y^2}{16} - \frac{x^2}{9} = 1 $, which is in the form $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $.

Determine the values of $ a^2 $ and $ b^2 $. From the equation, $ a^2 = 16 $ and $ b^2 = 9 $. Therefore, $ a = 4 $ and $ b = 3 $.

Recall that for a hyperbola of the form $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $, the equations of the asymptotes are given by $ y = \pm \frac{a}{b}x $.

Substitute the values of $ a $ and $ b $ into the asymptote formula. This gives $ y = \pm \frac{4}{3}x $.

Conclude that the equations for the asymptotes of the hyperbola are $ y = \pm \frac{4}{3}x $.

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