Question

Find standard form of the equation of the hyperbola that satisfies the given conditions:Vertices: (0, -2) and (0, 2)Asymptote: y=2x

Answer 1

aAs given by the question

There are given that the vertices and asymptote:

`\begin{gathered} \text{Vertices: (0, -2) and (0, 2)} \\ \text{Asymptote: y=2x} \end{gathered}`

Now,

First find the center (h, k):

So,

The center is the midpoint point of the vertices:

Then,

`\begin{gathered} (h,\text{ k)=(}(0+0)/(2),\text{ }(-2+2)/(2)) \\ (h,\text{ k)=(0},\text{ 0}) \end{gathered}`

Now,

Find the value of a

So,

a is the distance of center to vertices

So,

`\begin{gathered} a=\sqrt[]{(0+2)^2+(0-0)^2} \\ a=\sqrt[]{4} \\ a=2 \\ a^2=4 \end{gathered}`

Now,

From from the slope:

`(a)/(b)`

Then,

From the given asymptote, the slope of the equation is 2 which is given in the equation of asymptote.

So,

`\begin{gathered} (a)/(b)=2 \\ (2)/(b)=2 \\ 2b=2 \\ b=1 \end{gathered}`

Now,

From the standard form of the hyperbola:

`((y-k)^2)/(a^2)+((x-h)^2)/(b^2)=1`

Then,

`\begin{gathered} ((y-k)^2)/(a^2)+((x-h)^2)/(b^2)=1 \\ \frac{(y-0)^2}{4^{}}+\frac{(x-0)^2}{1^{}}=1 \end{gathered}`

Hence, the equation of hyperbola is shown below:

`\begin{gathered} \frac{(y-0)^2}{4^{}}+\frac{(x-0)^2}{1^{}}=1 \\ \frac{(y)^2}{4^{}}+\frac{(x)^2}{1^{}}=1 \end{gathered}`