Chapter 8.3, Problem 52E

Solution

An equation for the given hyperbola.

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

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 $\left(0,-\sqrt{2}\right)$ and $\left(0,\sqrt{2}\right)$ .

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

  $h=0$ , $k+a=\sqrt{2}$ and $k-a=-\sqrt{2}$ .

Adding the equations $k+a=\sqrt{2}$ and $k-a=-\sqrt{2}$ , it follows that,

  $2k=0$

Simplifying,

  $k=0$

Put $k=0$ in $k+a=\sqrt{2}$ to get $a=\sqrt{2}$ .

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

  $\frac{y^2}{2}-\frac{x^2}{b^2}=1$

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

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

  $\begin{array}{c}\frac{{\left(-2\right)}^2}{2}-\frac{2^2}{b^2}=1\\ \frac{4}{2}-\frac{4}{b^2}=1\\ 2-\frac{4}{b^2}=1\\ \frac{4}{b^2}=1\end{array}$

Solving,

  $b^2=4$

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

  $\frac{y^2}{2}-\frac{x^2}{4}=1$

This is the equation of the given hyperbola.

Conclusion:

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