Determine the vertices and foci of the hyperbola y24−x2=1\frac{y^2}{4}-x^2=14y2−x2=1.

Vertices: ( 0 , 2 ) , ( 0 , − 2 ) \left(0,2\right),\left(0,-2\right) ( 0 , 2 ) , ( 0 , − 2 ) Foci: ( 0 , 5 ) , ( 0 , − 5 ) \left(0,\sqrt5\right),\left(0,-\sqrt5\right) ( 0 , 5 <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"> ) , ( 0 , − 5 <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"> )

Determine the vertices and foci of the hyperbola y 2 4 − x 2 = 1 \frac{y^2}{4}-x^2=1 4 y 2 − x 2 = 1 .

Hey everyone. Let's see if we can solve this problem. So in this problem we're asked to determine the vertices and foci of the following hyperbola. Now to solve this, what I'm going to do is draw a graph of what I think the hyperbola is going to look like. And keep in mind, our goal is not to graph the hyperbola, our goal is to find the vertices and foci, but the graph can help us to see what this whole situation is gonna look like. Now notice how the y squared term comes first. Because the y squared term comes first in front of the x squared, that means we're going to have some kind of vertical hyperbola, which is something like that. Now, what we need to do is write the standard equation for the vertical hyperbola, which is going to be y squared over a squared minus x squared over b squared is equal to 1. And what this does is tell me what the a and b values are going to be related to in the original equation. Now I can see that a squared is going to be equivalent to what we have in the denominator of y squared, which is 4. So a squared is equal to 4, and to find a, I just take the square root on both sides, giving me a is the square root of 4, which is 2. Now we also need to find b squared, and b squared is going to be equivalent to whatever is in the denominator of x squared. But since we don't have anything in this denominator, we can just assume that there's a one underneath it, because x squared divided by 1 is just x squared. So b squared is going to be equal to 1, and if you take the square root on both sides here, we'll get that b equals 1. So these are the a and b values. Now with these a and b values, I can find c, which is the distance to the to the 2 foci. And recall that c squared for a hyperbola is equal to a squared plus b squared. Now we said that c a squared is gonna be 2 squared, which is 4, and b squared is gonna be 1 squared, which is 1. So I'll have c squared is equal to 4 plus 1, which is 5, and that take the square root of on both sides, we'll get that c is equal to 5, or excuse me, the square root of 5. So this right here is gonna be our a, b, and c values. Now from here, this allows me to find the vertices and foci, because recall that the vertices are going to be this distance to the closest point on the curve. So what that means is that our vertices are going to be at the points 0 2, and 0 negative 2, because this would be the point 2, and then that would be the point 0 negative 2 for the vertices, and then for the foci, the foci are going to be a distance C from the center. So we start at the center, we can go a distance C to get up to our first foci, and distance C down to our next foci. So our foci are going to be at 0 square root of 5 and 0 negative square root of 5. So this right here is going to be the vertices and the foci for the hyperbola we're dealing with. So that's how you can solve this problem. Hope you found this helpful. Thanks for watching.

Determine the vertices and foci of the hyperbola y 2 4 − x 2 = 1 \frac{y^2}{4}-x^2=1 4 y 2 − x 2 = 1 .

Vertices: ( 0 , 2 ) , ( 0 , − 2 ) \left(0,2\right),\left(0,-2\right) ( 0 , 2 ) , ( 0 , − 2 ) Foci: ( 0 , 5 ) , ( 0 , − 5 ) \left(0,\sqrt5\right),\left(0,-\sqrt5\right) ( 0 , 5 <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"> ) , ( 0 , − 5 <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"> )

Vertices: ( 2 , 0 ) , ( − 2 , 0 ) \left(2,0\right),\left(-2,0\right) ( 2 , 0 ) , ( − 2 , 0 ) Foci: ( 5 , 0 ) , ( − 5 , 0 ) \left(\sqrt5,0\right),\left(-\sqrt5,0\right) ( 5 <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"> , 0 ) , ( − 5 <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"> , 0 )

Vertices: ( 1 , 0 ) , ( − 1 , 0 ) \left(1,0\right),\left(-1,0\right) ( 1 , 0 ) , ( − 1 , 0 ) Foci: ( 5 , 0 ) , ( − 5 , 0 ) \left(5,0\right),\left(-5,0\right) ( 5 , 0 ) , ( − 5 , 0 )

Vertices: ( 0 , 1 ) , ( 0 , − 1 ) \left(0,1\right),\left(0,-1\right) ( 0 , 1 ) , ( 0 , − 1 ) Foci: ( 0 , 5 ) , ( 0 , − 5 ) \left(0,5\right),\left(0,-5\right) ( 0 , 5 ) , ( 0 , − 5 )

8. Conic Sections

Hyperbolas at the Origin

College Algebra

8. Conic Sections

Hyperbolas at the Origin

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Multiple Choice

Determine the vertices and foci of the hyperbola $\frac{y^2}{4}-x^2=1$ .

Vertices: $\left(2,0\right),\left(-2,0\right)$

Foci: $\left(\sqrt5,0\right),\left(-\sqrt5,0\right)$

Vertices: $\left(0,2\right),\left(0,-2\right)$

Foci: $\left(0,\sqrt5\right),\left(0,-\sqrt5\right)$

Vertices: $\left(1,0\right),\left(-1,0\right)$

Foci: $\left(5,0\right),\left(-5,0\right)$

Vertices: $\left(0,1\right),\left(0,-1\right)$

Foci: $\left(0,5\right),\left(0,-5\right)$

Verified step by step guidance

Identify the standard form of the hyperbola equation. The given equation is $ \frac{y^2}{4} - x^2 = 1 $, which is in the form $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $. This indicates a vertical hyperbola.

Determine the values of $ a^2 $ and $ b^2 $. From the equation, $ a^2 = 4 $ and $ b^2 = 1 $. Therefore, $ a = 2 $ and $ b = 1 $.

Find the vertices of the hyperbola. For a vertical hyperbola, the vertices are located at $ (0, \pm a) $. Thus, the vertices are $ (0, 2) $ and $ (0, -2) $.

Calculate the distance to the foci using the formula $ c^2 = a^2 + b^2 $. Substitute the known values: $ c^2 = 4 + 1 = 5 $, so $ c = \sqrt{5} $.

Determine the coordinates of the foci. For a vertical hyperbola, the foci are at $ (0, \pm c) $. Therefore, the foci are $ (0, \sqrt{5}) $ and $ (0, -\sqrt{5}) $.

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