Question

Which standard form of the equation of the hyperbola has vertices at (0, 5) and (0, –5), and asymptotes y equals plus or minus five twelfths times x question mark

y squared over 25 minus x squared over 144 equals 1
y squared over 144 minus x squared over 25 equals 1
x squared over 25 minus y squared over 144 equals 1
x squared over 144 minus y squared over 25 equals 1

Answer 1

The equation first represents the hyperbola has vertices at (0, 5) and (0, –5), and asymptotes y = ±(5/12)x option first is correct.

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:

Vertices of the hyperbola = (0, 5) and (0, -5)

Asymptotes: y = ±(5/12)x

The equations we have:

$\rm \dfrac{y^{2}}{25}-\dfrac{x^{2}}{144}=1$

$\rm \dfrac{y^{2}}{144}-\dfrac{x^{2}}{25}=1$

$\rm \dfrac{x^{2}}{25}-\dfrac{y^{2}}{144}=1$

$\rm \dfrac{y^{2}}{144}-\dfrac{x^{2}}{25}=1$

From the equation first:

$\rm \dfrac{y^{2}}{25}-\dfrac{x^{2}}{144}=1$

The value of a and b are:

a = 12

b = 5

Vertices of the hyperbola = (0, b) and (0, -b)

Vertices of the hyperbola = (0, 5) and (0, -5)

Asymptotes: y = ±(b/a)x

Asymptotes: y = ±(5/12)x

Thus, the equation first represents the hyperbola has vertices at (0, 5) and (0, –5), and asymptotes y = ±(5/12)x option first is correct.

Learn more about the hyperbola here:

brainly.com/question/12919612

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