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 1. 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 that 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 they have a horizontal hyperbola over a squared minus y minus k squared over b squared is equal to 1. And what I notice for this equation is that the h is going to correspond with this 2, but since we have a positive 2 and it's minus h, we need to make that number negative. So the h is going to be negative 2, and then the k corresponds with that 4, and there's a minus sign for both. So that's good. So k is equal to 4, meaning that our center is going to be at the position negative two 4. So graphically, the center would be back 2 units or actually no it would be yeah. It would be back 2 units and it would be up 1, 2, 3, 4 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 9, the first thing shows up in the denominator is 9. So a squared is equal to 9, and then we take the square root on both sides, we'll get the a is the square root of 9, which is 3. Now since we are centered at the right right about there, and the a value is 3, what we can do is go to the left and right 3 since we're dealing with a horizontal hyperbola. Now if I go to the left 1, 2, 3 units, it's gonna put one of our vertices here. And that's going to be at the position 1, 2, 3, 4, negative 5. This is right here at negative 5, and then vertically we are at a position of 4. All I can do is start from the center and I can go forward 3 units so we're going to go 1, 2, 3 units 4, which is going to put us at the position of horizontally we're at 1, and then vertically we're still at 4. So our vertices are going to be at negative 54 and at 14. 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 going to 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 9, and then b squared will be this whole second term, which is 16. That means c squared is 9 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 5. So So we're gonna go 5 units to the left, and we're gonna go 5 units to the right, which put us at 1, 2, 3, 4, 5 right about there. So what I can do is going from our center, if we're going back 5, well, one of our foci is going to be at negative 74, and if we go to the right 5, one of our foci is gonna be at 3 4, and these are going to be the foci. So the center is at negative 24, we're dealing with a horizontal hyperbola, our vertices are negative 54 and 14, and then our foci are negative 7434. So that is how you can find the vertices, the foci, the 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)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).

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).

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 $$\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 standard form of a hyperbola equation: $ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 $ for a horizontal hyperbola, and $ \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 standard form. Notice that the $ x $ term is positive and comes first, indicating a horizontal hyperbola.

Determine the center of the hyperbola from the equation $ (x+2)^2 $ and $ (y-4)^2 $. The center is $ (-2, 4) $.

Identify the vertices of the hyperbola. For a horizontal hyperbola, the vertices are $ (h \pm a, k) $. Here, $ a^2 = 9 $, so $ a = 3 $. The vertices are $ (-2+3, 4) $ and $ (-2-3, 4) $, which are $ (1, 4) $ and $ (-5, 4) $.

Calculate the foci of the hyperbola. The distance to the foci $ c $ is found using $ c^2 = a^2 + b^2 $. Here, $ b^2 = 16 $, so $ c^2 = 9 + 16 = 25 $, giving $ c = 5 $. The foci are $ (-2+5, 4) $ and $ (-2-5, 4) $, which are $ (3, 4) $ and $ (-7, 4) $.

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Struggling with Precalculus?

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.

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Hyperbolas NOT at the Origin practice set

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