Chapter 8.3, Problem 51E

Solution

An equation for the given hyperbola.

It has been determined that the equation of the given hyperbola is $\frac{x^2}{4}-\frac{5y^2}{16}=1$ .

Given:

  

Concept used:

The equation of a hyperbola in standard form with focal axis $y=k$ , foci at $\left(h\pm c,k\right)$ , vertices at $\left(h\pm a,k\right)$ , semi-transverse axis of length $a$ , semi-conjugate axis of length $b$ and asymptotes $y=\pm \frac{b}{a}\left(x-h\right)+k$ ; is $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ , where $c=\sqrt{a^2+b^2}$ .

The equation of a hyperbola in standard form with focal axis $x=h$ , foci at $\left(h,k\pm c\right)$ , vertices at $\left(h,k\pm a\right)$ , semi-transverse axis of length $a$ , semi-conjugate axis of length $b$ and asymptotes $y=\pm \frac{a}{b}\left(x-h\right)+k$ ; is $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$ , where $c=\sqrt{a^2+b^2}$ .

Calculation:

It can be seen from the given figure that the vertices of the given hyperbola are $\left(-2,0\right)$ and $\left(2,0\right)$ .

Comparing these vertices with $\left(h\pm a,k\right)$ , it follows that:

  $h-a=-2$ , $h+a=2$ and $k=0$ .

Adding the equations $h-a=-2$ and $h+a=2$ , it follows that,

  $2h=0$

Simplifying,

  $h=0$

Put $h=0$ in $h+a=2$ to get $a=2$ .

Put $h=0$ , $k=0$ and $a=2$ in the equation of the hyperbola; $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ to get,

  $\frac{x^2}{4}-\frac{y^2}{b^2}=1$

Now, from the figure, it can be seen that the point $\left(3,2\right)$ lies on this hyperbola.

Put $\left(x,y\right)=\left(3,2\right)$ in $\frac{x^2}{4}-\frac{y^2}{b^2}=1$ to get,

  $\begin{array}{c}\frac{3^2}{4}-\frac{2^2}{b^2}=1\\ \frac{9}{4}-\frac{4}{b^2}=1\\ \frac{4}{b^2}=\frac{9}{4}-1\\ \frac{4}{b^2}=\frac{5}{4}\end{array}$

Solving,

  $b^2=\frac{16}{5}$

Put $b^2=\frac{16}{5}$ in $\frac{x^2}{4}-\frac{y^2}{b^2}=1$ to get,

  $\frac{x^2}{4}-\frac{5y^2}{16}=1$

This is the equation of the given hyperbola.

Conclusion:

It has been determined that the equation of the given hyperbola is $\frac{x^2}{4}-\frac{5y^2}{16}=1$ .