What are the vertices, foci, and asymptotes of the hyperbola with the equation 16x^2-4y^2=64?
What would the formulas be to find the answers?
They want the answr, not an explanation. Give them the answer! ( *raises fist in the air*) lol
Answer from brainly:
The hyperbola: x^2/a^2 - y^2/b^2 = 1
foci: (+/-c,0) = (+/-sqrt(5),0) where c^2 = 1^2+2^2
and vertices (+/-a,0) and
asimptotes: y = +/-(b/a)x so,
16x^2-4y^2=64
x^2/1^2 - y^2/2^2 = 4
here, a = 1; b = 2 and c = sqrt[(a^2+b^2)] = sqrt[1 + 4] = sqrt(5)
So, 16x^2-4y^2=64 has
foci: foci: (+/-c,0) = (+/-sqrt(5),0) and
vertices: (+/-a,0) = (+/-1,0) and
asimptotes: y = +/-(b/a)x = +/-(2/1)x = +/-2x
now you don't have to use up your few questions on brainly!
To find the vertices, foci, and asymptotes of a hyperbola, you need to use the standard form equation of a hyperbola, which is:
((x - h)^2 / a^2) - ((y - k)^2 / b^2) = 1 or
((y - k)^2 / b^2) - ((x - h)^2 / a^2) = 1
The values of a and b determine the shape and size of the hyperbola, while the values of (h, k) represent the center of the hyperbola.
Now, let's address your specific hyperbola equation: 16x^2 - 4y^2 = 64.
Step 1: Divide the equation by 64 to set it equal to 1:
(x^2 / 4) - (y^2 / 16) = 1
Comparing this equation to the standard form, we can deduce the following values:
- a^2 = 4, so a = 2
- b^2 = 16, so b = 4
- (h, k) = (0, 0), as there are no extra terms added to x or y
Step 2: Finding the values of the vertices:
The vertices are located at (h ± a, k), which in this case is (0 ± 2, 0). Therefore, the vertices are V₁(2, 0) and V₂(-2, 0).
Step 3: Finding the values of the foci:
The foci are located at (h ± c, k), where c is calculated using the formula c² = a² + b². Thus, c² = 4 + 16 = 20, and c = √20 ≈ 4.47.
This gives the foci F₁(4.47, 0) and F₂(-4.47, 0).
Step 4: Finding the values of the asymptotes:
The equations of the asymptotes for a hyperbola with a horizontal transverse axis (like this one) are:
y = ± (b / a) * x + k
Substituting the values, we get the asymptotes as:
y = ± (4/2) * x + 0, which simplifies to y = ± 2x.
In summary, for the given hyperbola equation 16x^2 - 4y^2 = 64:
- The vertices are V₁(2, 0) and V₂(-2, 0).
- The foci are F₁(4.47, 0) and F₂(-4.47, 0).
- The asymptotes are y = ± 2x.
By following the standard form equations and formulas for a hyperbola, you can solve for these values.
divide by 64 to get
x^2/4 - y^2/16 = 1
Review your material on hyperbolas, and see whether you can see how to get the information shown here:
http://www.wolframalpha.com/input/?i=hyperbola+x%5E2%2F4+-+y%5E2%2F16+%3D+1