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,

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

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:

$\frac{a}{b}$

Then,

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

So,

$\begin{gathered} \frac{a}{b}=2 \\ \frac{2}{b}=2 \\ 2b=2 \\ b=1 \end{gathered}$

Now,

From the standard form of the hyperbola:

$\frac{(y-k)^2}{a^2}+\frac{(x-h)^2}{b^2}=1$

Then,

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

Hence, the equation of hyperbola is shown below:

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