Chapter 8.3, Problem 73E

Solution

To Prove: The given hyperbolas have the same asymptotes and the conjugate axis of each of these hyperbolas is the transverse axis of the other hyperbola.

It has been shown that the given hyperbolas have the same asymptotes and the conjugate axis of each of these hyperbolas is the transverse axis of the other hyperbola.

Given:

The hyperbolas; $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ and $\frac{{\left(y-k\right)}^2}{b^2}-\frac{{\left(x-h\right)}^2}{a^2}=1$ .

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:

The given hyperbola is $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ .

As discussed, the asymptotes of this hyperbola are $y=\pm \frac{b}{a}\left(x-h\right)+k$ .

The transverse axis of this hyperbola is the line segment joining the vertices $\left(h+a,k\right)$ and $\left(h-a,k\right)$ . Equivalently, the transverse axis of this hyperbola is the line segment lying on the line $y=k$ , passing through $\left(h,k\right)$ and having length $2a$ .

The conjugate axis is the line segment lying on the line $x=h$ , passing through $\left(h,k\right)$ and having length $2b$ .

Similarly, the given hyperbola is $\frac{{\left(y-k\right)}^2}{b^2}-\frac{{\left(x-h\right)}^2}{a^2}=1$ .

Now, the given hyperbola is obtained by switching $a$ and $b$ in $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$ .

As discussed, for the hyperbola $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$ :

The asymptotes are $y=\pm \frac{a}{b}\left(x-h\right)+k$ .

The transverse axis of this hyperbola is the line segment joining the vertices $\left(h,k+a\right)$ and $\left(h,k-a\right)$ . Equivalently, the transverse axis of this hyperbola is the line segment lying on the line $x=h$ , passing through $\left(h,k\right)$ and having length $2a$ .

The conjugate axis is the line segment lying on the line $y=k$ , passing through $\left(h,k\right)$ and having length $2b$ .

Now, switching $a$ and $b$ to obtain the above for the given hyperbola $\frac{{\left(y-k\right)}^2}{b^2}-\frac{{\left(x-h\right)}^2}{a^2}=1$ :

The asymptotes are $y=\pm \frac{b}{a}\left(x-h\right)+k$ .

The transverse axis of this hyperbola is the line segment joining the vertices $\left(h,k+b\right)$ and $\left(h,k-b\right)$ . Equivalently, the transverse axis of this hyperbola is the line segment lying on the line $x=h$ , passing through $\left(h,k\right)$ and having length $2b$ .

The conjugate axis is the line segment lying on the line $y=k$ , passing through $\left(h,k\right)$ and having length $2a$ .

So, it can be seen that:

The asymptotes of both given hyperbolas are $y=\pm \frac{b}{a}\left(x-h\right)+k$ .

The transverse axis of $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ and the conjugate axis of $\frac{{\left(y-k\right)}^2}{b^2}-\frac{{\left(x-h\right)}^2}{a^2}=1$ is the line segment lying on the line $y=k$ , passing through $\left(h,k\right)$ and having length $2a$ .

The conjugate axis of $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ and the transverse axis of $\frac{{\left(y-k\right)}^2}{b^2}-\frac{{\left(x-h\right)}^2}{a^2}=1$ is the line segment lying on the line $x=h$ , passing through $\left(h,k\right)$ and having length $2b$ .

This proves that the given hyperbolas have the same asymptotes and the conjugate axis of each of these hyperbolas is the transverse axis of the other hyperbola.

Conclusion:

It has been shown that the given hyperbolas have the same asymptotes and the conjugate axis of each of these hyperbolas is the transverse axis of the other hyperbola.