Find the equation for a hyperbola with a center at (0,0)\left(0,0\right)(0,0), focus at (0,−6)\left(0,-6\right)(0,−6) and vertex at (0,4)\left(0,4\right)(0,4).

y 2 16 − x 2 20 = 1 \frac{y^2}{16}-\frac{x^2}{20}=1 16 y 2 − 20 x 2 = 1

Find the equation for a hyperbola with a center at ( 0 , 0 ) \left(0,0\right) ( 0 , 0 ) , focus at ( 0 , − 6 ) \left(0,-6\right) ( 0 , − 6 ) and vertex at ( 0 , 4 ) \left(0,4\right) ( 0 , 4 ) .

Hey, everyone. So let's give this problem a try. So we are asking this problem to find the equation for a hyperbola with a center at 0, a focus at 0 negative 6, and a vertex at 4. So we're given all the details of this hyperbola, and we're now asked to work backwards and find the equation for it. Now before I do anything, what I'm actually going to do is write out a graph. Because what I can do with this graph is find about what this hyperbola is going to look like, and predict what the equation is going to be from this graph. So now that I have the graph written let's see what we've got. Now I see that we have a focus at 0 negative 6. So what that means and we're centered at 0, so if our center is here, then our focus is going to be somewhere down here 1, 2, 3, 4, 5, 6 units. This right here would be the focus 0, negative 6. Now we also have a vertex at 4. So what that means is if we go up 1, 2, 3, 4 units, we're going to have the vertex be right about here at 4. This is the vertex, and then this is the focus, and this actually tells me a lot about the hyperbola already. What this tells me is that we're going to have a hyperbola with vertices at 0, 4, and 0, negative 4 because we have to have evenness with the hyperbola, and we're going to have our foci at 0, negative 6, and then up to more units at 0 positive 6. This would be our other focus point. And that also tells me that we must have a hyperbola which is vertical since it opens this direction and that direction. So that's what we have so far. Now from this point, what I can do is recognize since we have a vertical hyperbola, that the equation is going to be y squared first over a squared equal or excuse me, minus x squared over b squared is equal to 1. And from here I just need to figure out what a and b are to solve this problem. Well to find a, it's pretty straightforward because a is the distance to the 2 vertices. Since it's 4 units to get up and 4 units to get down to the 2 vertices, that means a is equal to 4. Now finding b is a little bit more tricky since we don't have the b values given to us, but what I can do is use the foci and the vertices to find this. Recall that for the foci the distance is 'c'. So this would be a c distance to get up here, and a c distance to get down there. And c squared is equal to a squared plus b squared for a hyperbola. Now we have that c is going to be 6 units up or 6 units down, so c would be 6, we have that a is going to be 4 we already discussed that right here so we have 4 squared and this is all plus b squared and at this point all I need to do is solve for b. So what I can do is recognize that 6 squared is 36, 4 squared is 16, and this is plus b squared. I can subtract 16 on both sides of this equation giving us that b squared is equal to 36 minuteus 16, which is 20. If I take the square root on both sides that will give me that b is equal to the square root of 20. So what I can do is plug the a and the b into the equation we have up here. So we'll have y squared over and then this is 44 right there, so 4 squared. So y squared over 4 squared minus x squared over square root of 20 squared is equal to 1. And 4 squared is 16, and then the square and the square root will cancel giving me y squared over 16, minus x squared over 20 is equal to 1. And this right here is the equation for the hyperbola. So notice how we were able to take the center focus and vertex, and we were able to work backwards to find this equation. So that's how you can solve these types of problems. Hope you found this helpful.

Find the equation for a hyperbola with a center at ( 0 , 0 ) \left(0,0\right) ( 0 , 0 ) , focus at ( 0 , − 6 ) \left(0,-6\right) ( 0 , − 6 ) and vertex at ( 0 , 4 ) \left(0,4\right) ( 0 , 4 ) .

y 2 16 − x 2 20 = 1 \frac{y^2}{16}-\frac{x^2}{20}=1 16 y 2 − 20 x 2 = 1

y 2 20 − × 2 16 = 1 \frac{y^2}{20}-\times\frac{^2}{16}=1 20 y 2 − × 16 2 = 1

y 2 4 − x 2 20 = 1 \frac{y^2}{4}-\frac{x^2}{\sqrt{20}}=1 4 y 2 − 20 <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"> x 2 = 1

y 2 20 − x 2 4 = 1 \frac{y^2}{\sqrt{20}}-\frac{x2}{4}=1 20 <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"> y 2 − 4 x 2 = 1

19. Conic Sections

Hyperbolas at the Origin

Precalculus

19. Conic Sections

Hyperbolas at the Origin

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Multiple Choice

Find the equation for a hyperbola with a center at $\left(0,0\right)$ , focus at $\left(0,-6\right)$ and vertex at $\left(0,4\right)$ .

$\frac{y^2}{16}-\frac{x^2}{20}=1$

$\frac{y^2}{20}-\times\frac{^2}{16}=1$

$\frac{y^2}{4}-\frac{x^2}{\sqrt{20}}=1$

$\frac{y^2}{\sqrt{20}}-\frac{x2}{4}=1$

Verified step by step guidance

Identify the standard form of the equation for a hyperbola centered at the origin. Since the focus and vertex are on the y-axis, the equation will be of the form $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $.

Determine the distance from the center to the vertex, which is $ a $. Given the vertex at $ (0, 4) $, we have $ a = 4 $. Therefore, $ a^2 = 16 $.

Determine the distance from the center to the focus, which is $ c $. Given the focus at $ (0, -6) $, we have $ c = 6 $.

Use the relationship $ c^2 = a^2 + b^2 $ to find $ b^2 $. Substitute $ c = 6 $ and $ a^2 = 16 $ into the equation: $ 6^2 = 16 + b^2 $.

Solve for $ b^2 $ to complete the equation. Once $ b^2 $ is found, substitute $ a^2 $ and $ b^2 $ back into the standard form $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $ to get the equation of the hyperbola.

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