This question is a part of a group with common instructions.
View group »
After some more algebra, and then moving all terms with [math]x[/math] or [math]y[/math] to the left side of the equation (and all others to the right), the resulting equation is:
[math] a^2 x^2  c^2 x^2 + a^2 y^2 [/math]  [math] = [/math]  [math] a^4 a^2 c^2[/math] 
[math](a^2  c^2)x^2 + a^2 y^2 [/math]  [math] = [/math]  [math] (a^2  c^2) a^2 [/math] 
[math] b^2 x^2 + a^2 y^2 [/math]  [math] = [/math]  [math] a^2 b^2 [/math] 
[math]x^2 / a^2 + y^2 / b^2 [/math]  [math] = [/math]  [math] 1 [/math] 
This is the formula for the equation of an ellipse centered at [math](0,0)[/math]. Why can the substitution [math]b^2 = a^2c^2[/math] be made?

Since [math]b[/math] is undefined so far, it can be defined as any value. Then, using the Pythagorean theorem, [math]a^2 + b^2 = c^2[/math], simply rearrange to solve for [math]b^2[/math].

Knowing that the semiminor axis is [math]b[/math] units long, one can substitute the square of this value for [math]a^2c^2[/math].

Since [math]b[/math] is not yet defined, it can be used to simplify the equation by defining [math]b^2 = a^2  c^2[/math]. A positive value for [math]a^2c^2[/math] exists since [math]a > c[/math].

Projecting [math]b[/math], the length of the semiminor axis, onto the semimajor axis it is seen that [math]b=ac[/math]. Then, simply square both sides of the equation.