Describe the hyperbola (x+2)29−(y−4)216=1\frac{\left(x+2\right)^2}{9}-\frac{\left(y-4\right)^2}{16}=19(x+2)2−16(y−4)2=1.

This is a horizontal hyperbola centered at ( − 2 , 4 ) \left(-2,4\right) ( − 2 , 4 ) with vertices at ( 1 , 4 ) , ( − 5 , 4 ) \left(1,4\right),\left(-5,4\right) ( 1 , 4 ) , ( − 5 , 4 ) and foci at ( 3 , 4 ) , ( − 7 , 4 ) \left(3,4\right),\left(-7,4\right) ( 3 , 4 ) , ( − 7 , 4 ) .

Describe the hyperbola ( x + 2 ) 2 9 − ( y − 4 ) 2 16 = 1 \frac{\left(x+2\right)^2}{9}-\frac{\left(y-4\right)^2}{16}=1 9 ( x + 2 ) 2 − 16 ( y − 4 ) 2 = 1 .

Welcome back, everyone. So let's give this problem a try. In this problem, we are asked to describe the hyperbola given by this equation. Now to describe this hyperbola, the first thing I'm going to do is determine the orientation of the hyperbola. Now what I notice is we have the standard form of a hyperbola because we have this fraction minus that fraction is equal to one. And the first variable that I see in this form is the x. And since I see the x variable first, that means we are dealing with a horizontal hyperbola. Now what I'm going to do is I'm going to draw a small graph of this hyperbola, and we don't actually need to draw a graph, but this is just gonna be for reference so we understand what's happening. So you can see we're gonna have some kind of horizontally oriented hyperbola. But what I now need to do is figure out what the center of this hyperbola is going to be. And I can use these numbers up here. Because recall that the standard form for a hyperbola is x minus h squared, assuming you have a horizontal hyperbola, over a squared minus y minus k squared over b squared is equal to one. And what I noticed for this equation is that the h is going to correspond with this two. But since we have a positive two and it's minus h, we need to make that number negative. So the h is going to be negative two, and then the k corresponds with that four, and there's a minus sign for both. So that's good. So k is equal to four, meaning that our center is going to be at the position negative two four. So graphically, the center would be back two units or actually, no. It would be yeah. It would be back two units, and it would be up one, two, three, four units. So the hyperbola is going to be something kind of like that. What I also see from this equation, another thing we can do is find the vertices. And to find the vertices, we need our a value. Now recall that a squared is associated with the first thing that shows up in the fraction. So we see that this nine, the first thing shows up in the denominator is nine. So a squared is equal to nine, and then we take the square root on both sides, we'll get the a is the square root of nine, which is three. Now since we are centered at the right right about there, and the a value is three, what we can do is go to the left and right three since we're dealing with a horizontal hyperbola. Now if I go to the left one, two, three units, it's gonna put one of our vertices here, and that's going to be at the position one, two, three, four, negative five. This is right here at negative five, and then vertically we are at a position of four. All I can do is start from the center and I can go forward three units. So we're going to go one, two, three units four which is going to put us at the position of horizontally we're at one, and then vertically we're still at four. So our vertices are going to be at negative five, four and at one, four. So those are the vertices. Now our last step is going to be to find the foci, and we'll draw what the curves are gonna look like. So and we don't necessarily need to actually graph these curves, but we have the the curves like this. We know that the foci are going to be somewhere outside of these curves, because that's how it typically ends up. So we'll have something right there and right there about. And the foci, the way we can calculate them is by finding c. And for a hyperbola, c squared is equal to a squared plus b squared. Now a squared is going to be this whole first term that we see in the denominator which is nine, and then b squared will be this whole second term which is 16. That means c squared is nine plus 16 which is 25, and I could take the square root on both sides of this equation. So c is going to be equal to the square root of 25 which is five. So So we're gonna go five units to the left, and we're gonna go five units to the right, which is put us at one, two, three, four, five right about there. So what I can do is going from our center, if we're going back five, well one of our foci is going to be at negative seven four, and if we go to the right five, one of our foci is gonna be at 34. And these are going to be the foci. So the center is at negative two four, we're dealing with a horizontal hyperbola, our vertices are negative five four and one four, and then our foci are negative seven four three four. So that is how you can find the vertices, the foci, center, and the orientation of the hyperbola, and that is how I would best describe this hyperbola specifically. So hope you found this video helpful. Thanks for watching.

Describe the hyperbola ( x + 2 ) 2 9 − ( y − 4 ) 2 16 = 1 \frac{\left(x+2\right)^2}{9}-\frac{\left(y-4\right)^2}{16}=1 9 ( x + 2 ) 2 − 16 ( y − 4 ) 2 = 1 .

This is a horizontal hyperbola centered at ( − 2 , 4 ) \left(-2,4\right) ( − 2 , 4 ) with vertices at ( 1 , 4 ) , ( − 5 , 4 ) \left(1,4\right),\left(-5,4\right) ( 1 , 4 ) , ( − 5 , 4 ) and foci at ( 3 , 4 ) , ( − 7 , 4 ) \left(3,4\right),\left(-7,4\right) ( 3 , 4 ) , ( − 7 , 4 ) .

This is a vertical hyperbola centered at ( − 2 , 4 ) \left(-2,4\right) ( − 2 , 4 ) with vertices at ( 4 , 2 ) , ( 4 , − 6 ) \left(4,2\right),\left(4,-6\right) ( 4 , 2 ) , ( 4 , − 6 ) and foci at ( 4 , 4 ) , ( 4 , − 8 ) \left(4,4\right),\left(4,-8\right) ( 4 , 4 ) , ( 4 , − 8 ) .

This is a vertical hyperbola centered at ( 2 , − 4 ) \left(2,-4\right) ( 2 , − 4 ) with vertices at ( 4 , 1 ) , ( 4 , − 5 ) \left(4,1\right),\left(4,-5\right) ( 4 , 1 ) , ( 4 , − 5 ) and foci at ( 4 , 3 ) , ( 4 , − 7 ) \left(4,3\right),\left(4,-7\right) ( 4 , 3 ) , ( 4 , − 7 ) .

This is a horizontal hyperbola centered at ( − 2 , 4 ) \left(-2,4\right) ( − 2 , 4 ) with vertices at ( 2 , 4 ) , ( − 6 , 4 ) \left(2,4\right),\left(-6,4\right) ( 2 , 4 ) , ( − 6 , 4 ) and foci at ( 4 , 4 ) , ( − 8 , 4 ) \left(4,4\right),\left(-8,4\right) ( 4 , 4 ) , ( − 8 , 4 ) .

16. Parametric Equations & Polar Coordinates

Conic Sections

Calculus

16. Parametric Equations & Polar Coordinates

Conic Sections

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Multiple Choice

Describe the hyperbola $\frac{\left(x+2\right)^2}{9}-\frac{\left(y-4\right)^2}{16}=1$ .

This is a vertical hyperbola centered at $\left(-2,4\right)$ with vertices at $\left(4,2\right),\left(4,-6\right)$ and foci at $\left(4,4\right),\left(4,-8\right)$ .

This is a vertical hyperbola centered at $\left(2,-4\right)$ with vertices at $\left(4,1\right),\left(4,-5\right)$ and foci at $\left(4,3\right),\left(4,-7\right)$ .

This is a horizontal hyperbola centered at $\left(-2,4\right)$ with vertices at $\left(2,4\right),\left(-6,4\right)$ and foci at $\left(4,4\right),\left(-8,4\right)$ .

This is a horizontal hyperbola centered at $\left(-2,4\right)$ with vertices at $\left(1,4\right),\left(-5,4\right)$ and foci at $\left(3,4\right),\left(-7,4\right)$ .

Verified step by step guidance

Identify the general form of the hyperbola equation: $ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 $ for a horizontal hyperbola, or $ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 $ for a vertical hyperbola.

Compare the given equation $ \frac{(x+2)^2}{9} - \frac{(y-4)^2}{16} = 1 $ with the general form. Notice that the $ x $-term comes first, indicating this is a horizontal hyperbola.

Determine the center of the hyperbola $ (h, k) $ by identifying the values of $ h $ and $ k $ from the equation. Here, $ h = -2 $ and $ k = 4 $, so the center is $ (-2, 4) $.

Find the vertices. For a horizontal hyperbola, the vertices are located $ a $ units to the left and right of the center. Here, $ a^2 = 9 $, so $ a = 3 $. The vertices are $ (-2+3, 4) = (1, 4) $ and $ (-2-3, 4) = (-5, 4) $.

Calculate the foci. The distance from the center to each focus is $ c $, where $ c^2 = a^2 + b^2 $. Here, $ a^2 = 9 $ and $ b^2 = 16 $, so $ c^2 = 9 + 16 = 25 $, and $ c = 5 $. The foci are $ (-2+5, 4) = (3, 4) $ and $ (-2-5, 4) = (-7, 4) $.

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