Chapter 9.3, Problem 99E

To determine

To prove: show that $c^2=a^2+b^2$ for the equation of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .

Answer to Problem 99E

The asymptotes are $y=\frac{3}{2}x+\frac{1}{2}$ and $y=-\frac{3}{2}x-\frac{1}{2}$ .

Explanation of Solution

Given information:

The distance from the centre of the hyperbola $\left(0,0\right)$ to a focus is $c$ .

Calculation:

Draw perpendicular from $F_2$ to the x axis, which intersect the hyperbola at $P=\left(c,y\right)$ .

Now, by the definition of hyperbola,

  $P{F}_1=y+2a$ , by Pythagoras theorem

  $\begin{array}{l}{\left(y+2a\right)}^2=y^2+{\left(2c\right)}^2\\ y^2+4ay+4a^2=y^2+4c^2\\ 4ay=y^2+4c^2-4a^2-y^2\\ 4ay=4c^2-4a^2\\ y=\frac{c^2-a^2}{a}\end{array}$

Thus, point $P=\left(c,\frac{c^2-a^2}{a}\right)$ .

Substitute this point in the given hyperbola and solve:

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

Hence, proved.