Chapter 8, Problem 46RE

Solution

To find: the equation for the conic in the standard form.

The standard equation of the hyperbola with the vertices $\left(\pm 2,0\right)$ is $\frac{x^2}{4}-\frac{y^2}{16}=1$ .

Given data:

A hyperbola is given with center $\left(0,0\right)$ ,vertices $\left(\pm 2,0\right)$ and asymptotes as $y=\pm 2x$ .

Formula used:

The standard equation of the hyperbola with center $\left(0,0\right)$ and x-axis as focal axis is given as:

  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Vertices of the ellipse is $\left(\pm a,0\right)$ .

Asymptotes : $y=\pm \frac{b}{a}x$

Calculation:

It is given that the vertices of the hyperbola is $\left(\pm 2,0\right)$ .

  $a=2$

The asymptote of the ellipse is $y=\pm 2x$ .

  $\begin{array}{c}\pm \frac{b}{a}x=\pm 2x\\ \frac{b}{2}=2\\ b=4\end{array}$

The standard equation of the hyperbola with center $\left(0,0\right)$ and x-axis as focal axis is given as:

  $\begin{array}{l}\frac{x^2}{2^2}-\frac{y^2}{4^2}=1\\ \frac{x^2}{4}-\frac{y^2}{16}=1\end{array}$

Therefore, the standard equation of the hyperbola with the vertices $\left(\pm 2,0\right)$ is $\frac{x^2}{4}-\frac{y^2}{16}=1$