Find the equations for the asymptotes of the hyperbola x264−y2100=1\frac{x^2}{64}-\frac{y^2}{100}=164x2−100y2=1.

y = ± 5 4 x y=\pm\frac54x y = ± 4 5 x

Find the equations for the asymptotes of the hyperbola x 2 64 − y 2 100 = 1 \frac{x^2}{64}-\frac{y^2}{100}=1 64 x 2 − 100 y 2 = 1 .

Welcome back, everyone. Let's give this problem a try. Through this problem, we're asked to find the equations for the asymptotes of this hyperbola right here. Now to solve this problem, what I'm first gonna notice is that we have an x squared that shows up first. Because the x squared shows up first, that means we have a horizontal hyperbola. Now what I'm going to do is draw a graph, and this is not actually required for this problem, but this is more just for reference. So since we have a horizontal hyperbola, the hyperbola is gonna look something like that. Now what I can do is recognize that recall from previous videos that we discussed, you can make a rectangular box, which will tell you where the asymptotes will be. So if our box looks something like this, then we can draw the asymptotes through the corners of this box. So I can draw one of the asymptotes right through there, and then the other asymptotes will cross through the other corners of this box. Now, this allows me to see what the asymptotes are gonna look like, and they also help us to figure out what the curves are going to be as well. Because they're gonna come very close to approaching the curve without actually touching it. So this would continue to extend with the asymptotes. And what this tells me is how I can determine what our equation is gonna look like, and therefore find the equations for each of these asymptotes. Because the equation for each asymptotes is gonna be y equals plus or minus the slope of each of these lines times x. And the slope can be found by realizing that we have a rise over a run. So to find the slope, we'll start here at the origin. And by the way, this is why we only have to worry about the mx term because we're at the it crosses through 0, so we don't have anything out here. This would just be plus 0. And what we can do is recognize that we have to rise a distance of b. This is gonna be our rise, and then we have to run a distance of a because that would be the same distance to the vertex right there on the hyperbola. So if our slope is rise over run, that means we're gonna have y is equal to plus or minus the rise, which is b divided by the run, which is a, all multiplied by x. And that's the equation that we use when dealing with a horizontal hyperbola. So just in case you were wondering how we could remember this equation, or how it related to the shape, that's what we can do. So to solve this problem, all I need to do is figure out what 'b' and 'a' are. Well, a squared' is always associated with the first number that we see in the denominator, and b squared is the second number for a hyperbola. So if a squared is equal to 64, that means that a is gonna be the square root of 64, which is 8. And b squared is going to be equal to this number, which is a 100. And if I take the square root on both sides here, we'll get that b is the square root of a 100 which is 10. So this is going to be our a and our b, and I just need to plug that into this equation. So we'll have that y is equal to plus or minus b, which we determined to be 10, divided by a, which we determined to be 8 multiplied by x. Now I can simplify this down further, because what I recognize is that 10 is the same thing as 2 times 5, and 8 is the same thing as 2 times 4, as I'll multiply by x. So you can cancel one of the 2's here, giving me that y is equal to plus or minus, and then we're going to have 5 over 4x. So the most simplified answer for the asymptotes of this hyperbola is y is equal to plus or minus 5 over 4x, where the positive 5 over 4x is going to be this line that slopes up, and the negative 5 over 4x is going to be the line that slopes down. So So this is how you can find the equation for the asymptotes. Hope you found this helpful. Thanks for watching.

Find the equations for the asymptotes of the hyperbola x 2 64 − y 2 100 = 1 \frac{x^2}{64}-\frac{y^2}{100}=1 64 x 2 − 100 y 2 = 1 .

y = ± 5 4 x y=\pm\frac54x y = ± 4 5 x

y = ± 4 5 x y=\pm\frac45x y = ± 5 4 x

y = ± 16 25 x y=\pm\frac{16}{25}x y = ± 25 16 x

y = ± 25 16 x y=\pm\frac{25}{16}x y = ± 16 25 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 $\frac{x^2}{64}-\frac{y^2}{100}=1$ .

$y=\pm\frac45x$

$y=\pm\frac54x$

$y=\pm\frac{16}{25}x$

$y=\pm\frac{25}{16}x$

Verified step by step guidance

Identify the standard form of the hyperbola equation. The given equation is $ \frac{x^2}{64} - \frac{y^2}{100} = 1 $, which is in the form $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $.

Determine the values of $ a $ and $ b $ from the equation. Here, $ a^2 = 64 $ and $ b^2 = 100 $, so $ a = 8 $ and $ b = 10 $.

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

Substitute the values of $ a $ and $ b $ into the asymptote formula. This gives $ y = \pm \frac{10}{8}x $.

Simplify the fraction $ \frac{10}{8} $ to $ \frac{5}{4} $. Therefore, the equations of the asymptotes are $ y = \pm \frac{5}{4}x $.

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