Chapter B, Problem 51E

To determine

To calculate: To use the definition of a hyperbola to derive the equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ , $c^2=a^2+b^2$ for a hyperbola with foci $\left(\pm c,0\right)$

Answer to Problem 51E

Equation of hyperbola is derived

Explanation of Solution

Given information: Equation of hyperbola is $x^2-y^2=1$

Formula Used:

Hyperbola is the set of all the points in the plane whose difference from two points (the foci) is constant

Distance between two points is given as

  $D=\sqrt{{\left(x_1+x_2\right)}^2+{\left(y_1+y_2\right)}^2}$

Calculation:

Let us consider the below figure

  

From above figure,

  $\begin{array}{l}{F}_{1}P=\sqrt{{\left(x+c\right)}^2+y^2}\\ F_{2}P=\sqrt{{\left(x-c\right)}^2+y^2}\end{array}$

Also, given that thesum of the distances from point $P\left(x,y\right)$ on the ellipse to the foci is $2a$

Thus,

  $\begin{array}{l}{F}_{1}P+F_{2}P=\pm 2a\\ \sqrt{{\left(x+c\right)}^2+y^2}+\sqrt{{\left(x-c\right)}^2+y^2}=\pm 2a\\ \sqrt{{\left(x+c\right)}^2+y^2}=\pm 2a-\sqrt{{\left(x-c\right)}^2+y^2}\end{array}$

Squaring both sides of equation,

  $\begin{array}{l}{\left(\sqrt{{\left(x+c\right)}^2+y^2}\right)}^2={\left(\pm 2a-\sqrt{{\left(x-c\right)}^2+y^2}\right)}^2\\ {\left(x+c\right)}^2+y^2={\left(2a\right)}^2+{\left(x-c\right)}^2+y^2\mp 4a\sqrt{{\left(x-c\right)}^2+y^2}\\ x^2+c^2+2xc+y^2=4a^2+x^2+c^2-2xc+y^2\mp 4a\sqrt{{\left(x-c\right)}^2+y^2}\\ 4xc=4a^2\mp 4a\sqrt{{\left(x-c\right)}^2+y^2}\\ xc=a^2\mp a\sqrt{{\left(x-c\right)}^2+y^2}\\ xc-a^2=\mp a\sqrt{{\left(x-c\right)}^2+y^2}\end{array}$

Again, squaring both sides of equation

  $\begin{array}{l}{\left(xc-a^2\right)}^2={\left(-a\sqrt{{\left(x-c\right)}^2+y^2}\right)}^2\\ c^2{x}^2+a^4-2a^{2}cx=a^2\left({\left(x-c\right)}^2+y^2\right)\\ c^2{x}^2+a^4-2a^{2}cx=a^2\left(x^2+c^2-2cx+y^2\right)\\ c^2{x}^2+a^4-2a^{2}cx=a^2{x}^2+a^2{c}^2-2a^{2}cx+a^2{y}^2\\ c^2{x}^2+a^4=a^2{x}^2+a^2{c}^2+a^2{y}^2\\ c^2{x}^2-a^2{x}^2=a^2{c}^2-a^4+a^2{y}^2\\ \left(c^2-a^2\right)x^2=\left(c^2-a^2\right)a^2+a^2{y}^2\\ b^2{x}^2=b^2{a}^2+a^2{y}^2\\ b^2{x}^2-a^2{y}^2=b^2{a}^2\\ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\end{array}$

, where $c^2=a^2+b^2$

Conclusion:

Hence, equation of hyperbola is derived