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 ) , ( 1 , − 1 ) \left(1,1\right),\left(1,-1\right) ( 1 , 1 ) , ( 1 , − 1 ) 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"> ) .

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 one. 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 zero. And the h is going to be what is subtracted from x, which means we're going to have one be our h value. So that means our center is going to be at position one zero. 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 one zero, 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 two 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 one, and if I take the square root, we'll get that a is one. So what I can do is go one unit up and one unit down. So one of our vertices is going to be at the position one, and then zero plus one is one, and then another vertex point is going to be a one, and zero minus one is negative one. 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 one. And we can just see here that a squared would be one. And then plus b squared would be what we have under this denominator, which is four. So c squared is equal to one plus four, which is five. And if I take the square root on both sides, we'll get that c is equal to the square root of five. And our foci, well, if we have the same hyperbola that were centered at one zero and we have the points at right there for our vertices, then 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 one and then we can go up by the square root of five units. So since we have zero, we just add the square root of five to that and then we're also going to have the focus point at one and then we subtract the square root of five giving us negative square root of five and that's gonna be the foci. So for this hyperbola, it's a vertically oriented hyperbola that centers at one zero, 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 ) , ( 1 , − 1 ) \left(1,1\right),\left(1,-1\right) ( 1 , 1 ) , ( 1 , − 1 ) 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"> ) .

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"> ) .

16. Parametric Equations & Polar Coordinates

Conic Sections

Calculus

16. Parametric Equations & Polar Coordinates

Conic Sections

Struggling with Calculus?

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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 $\left(1,\sqrt5\right),\left(1,-\sqrt5\right)$ .