Chapter 10.4, Problem 73E

To determine

Derive the standard form of the hyperbola

Answer to Problem 73E

The standard form of hyperbola as $\left[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\right]$ .

Explanation of Solution

Given:

A hyperbola with center as origin and the horizontal transverse axis is given.

Let the foci of the hyperbola be $(f,0)$ and $(-f,0)$

Let $(x,y)$ be a point in the plane that lies on the hyperbola.

Let $2a$ be a positive constant.

Using the definition of hyperbola that the difference between the distances of the point from the foci is a positive constant,

  $\begin{array}{cc}\sqrt{{(x+f)}^2+{(y-0)}^2}-\sqrt{{(x-f)}^2+{(y-0)}^2}& =2a\\ \sqrt{x^2+f^2+2xf+y^2}& =2a+\sqrt{x^2+f^2-2xf+y^2}\end{array}$

Squaring both sides,

  $\begin{array}{cc}{x}^2+f^2+2xf+y^2& =4a^2+x^2+f^2-2xf+y^2+4a\sqrt{{(x-f)}^2+y^2}\\ 4fx& =4a^2+4a\sqrt{{(x-f)}^2+y^2}\\ 4fx-4a^2& =4a\sqrt{{(x-f)}^2+y^2}\end{array}$

Dividing both sides by 4a.

  $\frac{fx}{a}-a=\sqrt{\left(x^2-2fx+f^2+y^2\right)}$

  $\frac{fx-a^2}{a}=\sqrt{\left(x^2-2fx+f^2+y^2\right)}$

Squaring both sides,

  $\frac{a^4-2a^{2}fx+f^2{x}^2}{a^2}=x^2-2fx+f^2+y^2$

  $a^4-2a^{2}fx+f^2{x}^2=a^2{x}^2-2a^{2}fx+a^2{f}^2+a^2{y}^2$

  $f^2{x}^2-a^2{x}^2-a^2{y}^2=a^2{f}^2-a^4$

  $\left(f^2-a^2\right)x^2-a^2{y}^2=a^2\left(c^2-a^2\right)$

Dividing both sides by $a^2\left(f^2-a^2\right)$

  $\frac{x^2}{a^2}-\frac{y^2}{\left(c^2-a^2\right)}=1$

Let $b^2=c^2-a^2$

  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Hence derived the standard form of hyperbola as $\left[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\right]$ .