Chapter 9.5, Problem 4OE

To determine

To calculate: To find the x- and y-intercepts of the hyperbola and the equation of its asymptotes and to tell on which of the coordinate axes its foci lie

Answer to Problem 4OE

x-intercept is $2$ and y-intercept is $4$ and foci lies on x-axis and equation of asymptotes are $y=\pm 2x$

Explanation of Solution

Given information: Equation of hyperbola is $4x^2-y^2=16$

Formula Used:

The hyperbola with center $\left(0,0\right)$ and foci $\left(-c,0\right)$ and $\left(c,0\right)$ and difference of focal radii $2a$ has the equation:

  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ , where $b^2=c^2-a^2$

Equation of the asymptotes are $y=\frac{b}{a}x$ and $y=-\frac{b}{a}x$

The hyperbola with center $\left(0,0\right)$ and foci $\left(0,-c\right)$ and $\left(0,c\right)$ and difference of focal radii $2a$ has the equation:

  $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ , where $b^2=c^2-a^2$

Equation of the asymptotes are $y=\frac{a}{b}x$ and $y=-\frac{a}{b}x$

Calculation:

Equation of hyperbola is given as follows:

  $4x^2-y^2=16$

Divide both sides of equation by $16$ :

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

Comparing the equation with standard form of the equation of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ :

  $\begin{array}{l}a=2\\ b=4\end{array}$

Foci is calculated as follows:

  $\begin{array}{l}{b}^2=c^2-a^2\\ c^2=a^2+b^2\\ c^2=2^2+4^2\\ c^2=4+16\\ c^2=20\\ c=\pm 2\sqrt{5}\end{array}$

Thus, Foci are $\left(2\sqrt{5},0\right)$ and $\left(-2\sqrt{5},0\right)$

Also, asymptotes are calculated as follows:

  $y=\frac{b}{a}x$ and $y=-\frac{b}{a}x$

Substituting the values:

  $y=\frac{4}{2}x$   and  $y=-\frac{4}{2}x$

  $y=2x$   and  $y=-2x$

Conclusion:

Hence, x-intercept is $2$ and y-intercept is $4$ and foci lies on x-axis and equation of asymptotes are $y=\pm 2x$