Chapter 8.3, Problem 74E

Solution

To Show: If $x=c$ , then $y=\pm \frac{b^2}{a}$ for the given hyperbola and to determine why it is reasonable to define the focal width of such hyperbolas to be $\frac{2b^2}{a}$ .

It has been shown that if $x=c$ , then $y=\pm \frac{b^2}{a}$ for the given hyperbola and it has been determined why it is reasonable to define the focal width of such hyperbolas to be $\frac{2b^2}{a}$ .

Given:

The hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .

Concept used:

The equation of a hyperbola in standard form with focal axis $y=0$ , foci at $\left(\pm c,0\right)$ , vertices at $\left(\pm a,0\right)$ , semi-transverse axis of length $a$ , semi-conjugate axis of length $b$ and asymptotes $y=\pm \frac{b}{a}x$ ; is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ , where $c=\sqrt{a^2+b^2}$ .

Calculation:

The given hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .

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

  $\frac{c^2}{a^2}-\frac{y^2}{b^2}=1$

Put $c=\sqrt{a^2+b^2}$ in the above equation to get,

  $\begin{array}{c}\frac{a^2+b^2}{a^2}-\frac{y^2}{b^2}=1\\ 1+\frac{b^2}{a^2}-\frac{y^2}{b^2}=1\\ \frac{b^2}{a^2}-\frac{y^2}{b^2}=0\\ \frac{b^2}{a^2}=\frac{y^2}{b^2}\end{array}$

Solving,

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

This shows that for the given hyperbola, if $x=c$ , then $y=\pm \frac{b^2}{a}$ .

Now, the focal axis of the given hyperbola is $y=0$ , that is, the $x$ axis.

According to the definition, the focal width of a hyperbola is the length of the chord of the hyperbola passing through the focus and perpendicular to the focal axis.

As shown previously, for the given hyperbola, if $x=c$ , then $y=\pm \frac{b^2}{a}$ .

Then, the points $\left(c,\frac{b^2}{a}\right)$ and $\left(c,-\frac{b^2}{a}\right)$ lie on the given hyperbola.

Consider the chord joining these two points.

Note that the midpoint of this chord is given as: $\left(\frac{c+c}{2},\frac{-\frac{b^2}{a}+\frac{b^2}{a}}{2}\right)=\left(c,0\right)$ , which is a focus of the given hyperbola.

Also note that the points $\left(c,\frac{b^2}{a}\right)$ and $\left(c,-\frac{b^2}{a}\right)$ lie on the line $x=c$ , which implies that the chord joining them lies on the line $x=c$ , which is perpendicular to the focal axis $y=0$ , that is, the $x$ axis.

Then, the length of this chord must be the focal width.

Now, the length of the chord joining $\left(c,\frac{b^2}{a}\right)$ and $\left(c,-\frac{b^2}{a}\right)$ must be $\frac{b^2}{a}-\left(-\frac{b^2}{a}\right)=\frac{2b^2}{a}$ .

This justifies why it is reasonable to define the focal width of such hyperbolas to be $\frac{2b^2}{a}$ .

Conclusion:

It has been shown that if $x=c$ , then $y=\pm \frac{b^2}{a}$ for the given hyperbola and it has been determined why it is reasonable to define the focal width of such hyperbolas to be $\frac{2b^2}{a}$ .