Given the hyperbola y2100−x2139=1\frac{y^2}{100}-\frac{x^2}{139}=1100y2−139x2=1, find the length of the aaa -axis and the bbb -axis.

a = 10 a=10 a = 10 , b = 139 b=\sqrt{139} b = 139 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

Given the hyperbola y 2 100 − x 2 139 = 1 \frac{y^2}{100}-\frac{x^2}{139}=1 100 y 2 − 139 x 2 = 1 , find the length of the a a a -axis and the b b b -axis.

Welcome back everyone. Let's give this problem a try. So here we are given the hyperbola y squared over a 100 minus x squared over 139 is equal to 1. We're asked to find the length of the a and b axis. Now to solve this problem, notice that it's the y squared term that comes first. So that means we're dealing with a vertically oriented hyperbola, and the equation for a vertical hyperbola is y squared over a squared minus x squared over b squared is equal to 1. Now what I can do here is recognize that a squared is equal to 100. So I have a squared equals a 100, and if I take the square root on both sides of this equation, we'll get that a is equal to the square root of a 100, which is 10. Now what I can also recognize is that b squared is equal to 139. And to solve for b, I can just take the square root on both sides of this equation, canceling the square, giving us that b is simply the square root of 139. This does not simplify down any farther. So this would be our b value, and that would be a. And that is how you can find the 2 axes, so that's the answer to the problem. Hope you found this video helpful, thanks for watching.

Given the hyperbola y 2 100 − x 2 139 = 1 \frac{y^2}{100}-\frac{x^2}{139}=1 100 y 2 − 139 x 2 = 1 , find the length of the a a a -axis and the b b b -axis.

a = 10 a=10 a = 10 , b = 139 b=\sqrt{139} b = 139 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

a = 100 a=100 a = 100 , b = 139 b=139 b = 139

a = 139 a=139 a = 139 , b = 100 b=100 b = 100

a = 139 a=\sqrt{139} a = 139 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , b = 10 b=10 b = 10

19. Conic Sections

Hyperbolas at the Origin

Precalculus

19. Conic Sections

Hyperbolas at the Origin

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Multiple Choice

Given the hyperbola $\frac{y^2}{100}-\frac{x^2}{139}=1$ , find the length of the $a$ -axis and the $b$ -axis.

$a=100$ , $b=139$

$a=139$ , $b=100$

$a=\sqrt{139}$ , $b=10$

$a=10$ , $b=\sqrt{139}$

Verified step by step guidance

Identify the standard form of the hyperbola equation: $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $. In this form, the hyperbola is vertical, meaning the transverse axis is along the y-axis.

Compare the given equation $ \frac{y^2}{100} - \frac{x^2}{139} = 1 $ with the standard form to identify $ a^2 = 100 $ and $ b^2 = 139 $.

Calculate the value of $ a $ by taking the square root of $ a^2 $: $ a = \sqrt{100} = 10 $. This represents the semi-major axis length along the y-axis.

Calculate the value of $ b $ by taking the square root of $ b^2 $: $ b = \sqrt{139} $. This represents the semi-minor axis length along the x-axis.

Conclude that the length of the $ a $-axis (transverse axis) is $ 2a = 20 $ and the length of the $ b $-axis (conjugate axis) is $ 2b = 2\sqrt{139} $.

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