On the equation of an ellipse in analytic geometry

Question

Intuitively, why is the equation by the ellipse given by $(x^2/a^2)+(y^2/b^2)=1$? Why does the addition of $1/a^2$ and $1/b^2$ shift the shape of the circle into that of a rectangle? Also, why is the expression always equals to one?

By the way, I'm sorry about the formatting above. I'm new to MathStackExchange and don't know how to type an equation.

Answer 1

Don't think of the parameters drawing an ellipse. Think instead of the curve $x'^2+y'^2=1$, a unit circle. We draw this in a stretched space, contracting $x$ by $a$ and $y$ by $b$ through the substitutions $x'$->$x/a$ and $y'$->$y/b$.

Answer 2

Intuitively, why is the equation by the ellipse given by $(x^2/a^2)+(y^2/b^2)=1$?

This was answered by Captain Morgan. The equation $$ x^2 + y^2 = 1 $$ for a circle around $(0,0)$ with radius $1$ is derived from the equation $$ d((x,y), (0,0)) = 1 $$ where $d$ is the Euclidean distance between a point $(x,y)$ and the origin $(0,0)$, which according to Pythagoras' theorem is $$ x^2 + y^2 = 1^2 $$

Why does the addition of $1/a^2$ and $1/b^2$ shift the shape of the circle into that of a rectangle?

You should provide an example formula because it is not clear where you add this to what.

Also, why is the expression always equals to one?

For a circle the constant at the right hand side is the square of the radius. For an ellipse this splits into two similar parameters, the semi axes $a$ and $b$. E.g. see a derivation here: link.

Answer 3

Suppose that $a > b > 0$.

Choose any point $(u,v)$ on the unit circle. That is $u^2+v^2=1$.

Let $P=(x,y)$ where $x=au$ and $y=bv$. Then $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$.

Note that the transformation $x \to ax$ stretches the left and right sides of the unit circle horizontally by a factor of $a$ while the transformation $y \to by$ stretches the upper and lower sides of the unit circle vertically by a factor of $b$, resulting in an oval shape.

But is this figure a true ellipse? The foci will be at the points $F_1 = (-c,0)$ and $F_2 = (c,0)$ where $c=\sqrt{a^2-b^2}$.

The arithmetic is horrendous, but we compute

\begin{align} PF_1+PF_2 &= \sqrt{(x + c)^2+y^2} + \sqrt{(x - c)^2+y^2} \\ &= \sqrt{(au+c)^2+b^2v^2}+\sqrt{(au-c)^2+b^2v^2} \\ &= \sqrt{(au+c)^2+(a^2-c^2)(1-u^2)} +\sqrt{(au-c)^2+(a^2-c^2)(1-u^2)} \\ &= \sqrt{a^2 + 2acu + c^2 u^2} + \sqrt{a^2 - 2acu + c^2 u^2}\\ &= (a+cu) + (a-cu) \\ &= 2a \end{align}

Which is exactly what you would expect for an ellipse.