Describe the hyperbola y2−(x−1)24=1y^2-\frac{\left(x-1\right)^2}{4}=1y2−4(x−1)2=1.

This is a vertical hyperbola centered at ( 1 , 0 ) \left(1,0\right) ( 1 , 0 ) with vertices at ( 1 , 1 ) \left(1,1\right) ( 1 , 1 ) , ( 1 , − 1 ) \left(1,-1\right) ( 1 , − 1 ) and foci at ( 1 , 5 ) \left(1,\sqrt5\right) ( 1 , 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"> ) , ( 1 , − 5 ) \left(1,-\sqrt5\right) ( 1 , − 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"> ) .

Describe the hyperbola y 2 − ( x − 1 ) 2 4 = 1 y^2-\frac{\left(x-1\right)^2}{4}=1 y 2 − 4 ( x − 1 ) 2 = 1 .

Welcome back everyone. So let's give this problem a try. In this problem we're asked to describe the hyperbola given by this equation. Now, to start, what I'm going to do is determine the orientation of the hyperbola. I notice that the first variable that we see is y. So since we see a y in front, that means that we are dealing with a vertical hyperbola. So that's the orientation. Now the next thing I'm going to do is try to find the center of the hyperbola. And for our center we need to find h and k. Now recall that our standard equation is going to be for a vertical hyperbola y minus k squared over a squared minus x minus h squared over b squared is equal to 1. And since this is our standard equation, well I noticed that we don't have a k attached to that y variable. So our k is going to be 0, and the h is going to be what is subtracted from x, which means we're going to have 1 be our h value. So that means our center is going to be at position 10. Now the next thing I'm going to do is find the vertices. And to find the vertices, what I need to do is go a units to the or actually a units up and down from our center because we're dealing with a vertical hyperbola. So if you imagine that we have some hyperbola and the center is at 10, which is right about there, If this is a vertical hyperbola, our vertices are going to be up and down by a distance of a. So to find the distance to the 2 vertices, well what I'm going to do is recognize that a squared is going to be associated with what's in the denominator for this first term, but since it's by itself we can just imagine there's a one underneath it. So a squared is equal to 1, and if I take the square root we'll get that a is 1. So what I can do is go 1 unit up and 1 unit down. So one of our vertices is going to be at position 1, and then 0 plus 1 is 1, and then another vertex point is going to be a 1, and 0 minus 1 is negative 1. So those are the vertices. Now the last thing we need to do is find the foci. And for the foci, well we're gonna need our c. And c squared is equal to a squared plus b squared. Now we can see up here that a squared is going to be this first thing that we see and we discussed that there's an invisible one in that denominator so our a squared is going to be 1, and we can just see here that a squared would be 1, and then plus b squared would be what we have under this denominator which is 4. So c squared is equal to 1+4 which is 5, and if I take the square root on both sides we'll get that c is equal to the square root of 5. And our foci, well if we have the same hyperbola that we're centered at 10, and we have the points at right there for our vertices, and our foci are going to be up here and down there. So writing the focus points what we can do is we can start from our same horizontal position of 1, and then we can go up by the square root of 5 units. So since we have 0 we just add the square root of 5 to that, and then what we're also going to have is a focus point at 1, and then we subtract the square root of 5, giving us negative square root 5, and that's gonna be the foci. So for this hyperbola it's a vertically oriented hyperbola this centers at 10 and those are the vertices and those are the foci. So hope you found this video helpful thanks for watching.

Describe the hyperbola y 2 − ( x − 1 ) 2 4 = 1 y^2-\frac{\left(x-1\right)^2}{4}=1 y 2 − 4 ( x − 1 ) 2 = 1 .

This is a vertical hyperbola centered at ( 1 , 0 ) \left(1,0\right) ( 1 , 0 ) with vertices at ( 1 , 1 ) \left(1,1\right) ( 1 , 1 ) , ( 1 , − 1 ) \left(1,-1\right) ( 1 , − 1 ) and foci at ( 1 , 5 ) \left(1,\sqrt5\right) ( 1 , 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"> ) , ( 1 , − 5 ) \left(1,-\sqrt5\right) ( 1 , − 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"> ) .

This is a vertical hyperbola centered at ( 1 , 0 ) \left(1,0\right) ( 1 , 0 ) with vertices at ( 1 , 2 ) , ( 1 , − 2 ) \left(1,2\right),\left(1,-2\right) ( 1 , 2 ) , ( 1 , − 2 ) and foci at ( 1 , 1 ) , ( 1 , − 1 ) \left(1,1\right),\left(1,-1\right) ( 1 , 1 ) , ( 1 , − 1 ) .

This is a horizontal hyperbola centered at ( − 1 , 0 ) \left(-1,0\right) ( − 1 , 0 ) with vertices at ( 0 , 0 ) , ( − 2 , 0 ) \left(0,0\right),\left(-2,0\right) ( 0 , 0 ) , ( − 2 , 0 ) and foci at ( 5 − 1 , 0 ) , ( − 5 − 1 , 0 ) \left(\sqrt5-1,0\right),\left(-\sqrt5-1,0\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"> − 1 , 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"> − 1 , 0 ) .

This is a horizontal hyperbola centered at ( 1 , 0 ) \left(1,0\right) ( 1 , 0 ) with vertices at ( 0 , 0 ) , ( − 2 , 0 ) \left(0,0\right),\left(-2,0\right) ( 0 , 0 ) , ( − 2 , 0 ) and foci at ( 1 , − 5 ) , ( 1 , − 5 ) \left(1,-\sqrt5\right),\left(1,-\sqrt5\right) ( 1 , − 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"> ) , ( 1 , − 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"> ) .

19. Conic Sections

Hyperbolas NOT at the Origin

Precalculus

19. Conic Sections

Hyperbolas NOT at the Origin

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Multiple Choice

Describe the hyperbola $y^2-\frac{\left(x-1\right)^2}{4}=1$ .

This is a vertical hyperbola centered at $\left(1,0\right)$ with vertices at $\left(1,1\right)$ , $\left(1,-1\right)$ and foci at $\left(1,\sqrt5\right)$ , $\left(1,-\sqrt5\right)$ .

This is a vertical hyperbola centered at $\left(1,0\right)$ with vertices at $\left(1,2\right),\left(1,-2\right)$ and foci at $\left(1,1\right),\left(1,-1\right)$ .

This is a horizontal hyperbola centered at $\left(-1,0\right)$ with vertices at $\left(0,0\right),\left(-2,0\right)$ and foci at $\left(\sqrt5-1,0\right),\left(-\sqrt5-1,0\right)$ .

This is a horizontal hyperbola centered at $\left(1,0\right)$ with vertices at $\left(0,0\right),\left(-2,0\right)$ and foci at $\left(1,-\sqrt5\right),\left(1,-\sqrt5\right)$ .

Verified step by step guidance

Identify the standard form of a hyperbola equation. The given equation is $ y^2 - \frac{(x-1)^2}{4} = 1 $, which can be compared to the standard form $ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 $ for a vertical hyperbola.

Determine the center of the hyperbola. In the equation $ y^2 - \frac{(x-1)^2}{4} = 1 $, the center $(h, k)$ is $(1, 0)$ because the equation can be rewritten as $ \frac{(y-0)^2}{1} - \frac{(x-1)^2}{4} = 1 $.

Identify the vertices of the hyperbola. For a vertical hyperbola, the vertices are located at $(h, k \pm a)$. Here, $a^2 = 1$, so $a = 1$. Therefore, the vertices are $(1, 0 \pm 1)$, which are $(1, 1)$ and $(1, -1)$.

Calculate the foci of the hyperbola. The foci are located at $(h, k \pm c)$, where $c^2 = a^2 + b^2$. Here, $b^2 = 4$, so $c^2 = 1 + 4 = 5$, giving $c = \sqrt{5}$. Thus, the foci are $(1, 0 \pm \sqrt{5})$, which are $(1, \sqrt{5})$ and $(1, -\sqrt{5})$.

Summarize the characteristics of the hyperbola. This is a vertical hyperbola centered at $(1, 0)$ with vertices at $(1, 1)$ and $(1, -1)$, and foci at $(1, \sqrt{5})$ and $(1, -\sqrt{5})$.

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