Question

What are the vertices, foci and asymptotes of the hyperbola with equation 16x^2-4y^2=64

Answer 1

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

Answer 2

The vertices of the hyperbola are (±2, 0), foci of the hyperbola are (±2√5, 0) and asymptotes are y = 2x and y = -2x.

What is hyperbola?

It's a two-dimensional geometry curve with two components that are both symmetric. In other words, the number of points in two-dimensional geometry that have a constant difference between them and two fixed points in the plane can be defined.

We have an equation for the hyperbola:

$\rm 16x^2-4y^2=64\\\\\rm \dfrac{x^2}{4}- \dfrac{y^2}{16}= 1\\\\\rm \dfrac{x^2}{2^2}- \dfrac{y^2}{4^2}= 1$

Here h = 0, k = 0, a = 2, and b = 4

c = √(2²+4²) = 2√5

The first vertex is:

$\rm \left(h - a, \ k\right) =(-2,0)$

The second vertex is:

$\rm \left(h + a, k\right) = (2,0)$

The first focus is:

$\rm \left(h - c, \ k\right) = (-2\sqrt{5}, \ 0)$

The second focus is:

$\rm \left(h + c, k\right) =(2 \sqrt{5} \ ,0)$

​

The asymptote is:

$\rm y = \pm \frac{b}{a} \left(x - h\right) + k \\\\\\\rm =\pm 2x$

Thus, the vertices of the hyperbola are (±2, 0), foci of the hyperbola are (±2√5, 0) and asymptotes are y = 2x and y = -2x.

Learn more about the hyperbola here:

brainly.com/question/12919612

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