# How to find a neat version of y when you have the equation of an ellipse

So, you have the equation of the ellipse but you need to completely isolate y. How would you go about doing this? Well, here is a fantastic example…

${ \left( \frac { x }{ a } \right) }^{ 2 }+{ \left( \frac { y }{ b } \right) }^{ 2 }=1\\ \\ \therefore \quad { \left( \frac { y }{ b } \right) }^{ 2 }=1-{ \left( \frac { x }{ a } \right) }^{ 2 }\\ \\ \therefore \quad \frac { { y }^{ 2 } }{ { b }^{ 2 } } =1-\frac { { x }^{ 2 } }{ { a }^{ 2 } } \\ \\ \therefore \quad { y }^{ 2 }={ b }^{ 2 }-\frac { { { b }^{ 2 }x }^{ 2 } }{ { a }^{ 2 } } \\ \\ \therefore \quad { y }^{ 2 }=\frac { { a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 } } -\frac { { b }^{ 2 }{ x }^{ 2 } }{ { a }^{ 2 } } \\ \\ \therefore \quad { y }^{ 2 }=\frac { { a }^{ 2 }{ b }^{ 2 }-{ b }^{ 2 }{ x }^{ 2 } }{ { a }^{ 2 } } \\ \\ \therefore \quad { y }^{ 2 }=\frac { { b }^{ 2 }\left( { a }^{ 2 }-{ x }^{ 2 } \right) }{ { a }^{ 2 } } \\ \\ \therefore \quad { y }^{ 2 }=\frac { { b }^{ 2 } }{ { a }^{ 2 } } \cdot \left( { a }^{ 2 }-{ x }^{ 2 } \right) \\ \\ \therefore \quad y=\sqrt { \frac { { b }^{ 2 } }{ { a }^{ 2 } } } \cdot \sqrt { { a }^{ 2 }-{ x }^{ 2 } } \\ \\ \therefore \quad y=\frac { b }{ a } \cdot \sqrt { { a }^{ 2 }-{ x }^{ 2 } }$

This will come in handy if you’re trying to derive the area of an ellipse from absolute scratch.

# Deriving the formula for an ellipse

In this post, I’ll be demonstrating how one can derive the formula for an ellipse from absolute scratch.

To derive the formula for an ellipse, what we must first do is create a diagram like the one below.

** Click on the image above to see it in full size.

Now, the first thing we’ve got to acknowledge here is that:

${ D }_{ 1 }+{ D }_{ 2 }=2a$

What we’re basically saying is that D_1 + D_2 is equal to the length from -a to a in the diagram above.

This formula can be understood by watching the video below…

These photographs can also help the formula sink into your mind…

Ellipse Image 1:

Ellipse Image 2:

Now, look at the diagram at the top of this page once again…

What you will notice is that:

${ \left( c+x \right) }^{ 2 }+{ y }^{ 2 }={ D }_{ 1 }^{ 2 }\\ \\ \therefore \quad { D }_{ 1 }^{ 2 }={ c }^{ 2 }+2cx+{ x }^{ 2 }+{ y }^{ 2 }\\ \\ \therefore \quad { D }_{ 1 }=\sqrt { { c }^{ 2 }+2cx+{ x }^{ 2 }+{ y }^{ 2 } } \\ \\ { \left( c-x \right) }^{ 2 }+{ y }^{ 2 }={ D }_{ 2 }^{ 2 }\\ \\ \therefore \quad { D }_{ 2 }^{ 2 }={ c }^{ 2 }-2cx+{ x }^{ 2 }+{ y }^{ 2 }\\ \\ \therefore \quad { D }_{ 2 }=\sqrt { { c }^{ 2 }-2cx+{ x }^{ 2 }+{ y }^{ 2 } }$

If this is the case, we can say that:

** Click on the image of the workings to see it in full size.

Alright, so far so good… Now, it turns out – if you look at the diagram at the top of this page carefully, you will discover that:

${ b }^{ 2 }+{ c }^{ 2 }={ a }^{ 2 }\\ \\ \therefore \quad { c }^{ 2 }={ a }^{ 2 }-{ b }^{ 2 }$

And this ultimately means that:

${ a }^{ 4 }+\left( { a }^{ 2 }-{ b }^{ 2 } \right) { x }^{ 2 }={ a }^{ 2 }\left( { a }^{ 2 }-{ b }^{ 2 } \right) +{ a }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }\\ \\ \therefore \quad { a }^{ 4 }+{ a }^{ 2 }{ x }^{ 2 }-{ b }^{ 2 }{ x }^{ 2 }={ a }^{ 4 }-{ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }\\ \\ -{ b }^{ 2 }{ x }^{ 2 }=-{ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }\\ \\ { b }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }={ a }^{ 2 }{ b }^{ 2 }\\ \\ \frac { { b }^{ 2 }{ x }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } +\frac { { a }^{ 2 }{ y }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } =\frac { { a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } \\ \\ \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\\ \\ { \left( \frac { x }{ a } \right) }^{ 2 }+{ \left( \frac { y }{ b } \right) }^{ 2 }=1$

The formula you see just above is the formula for an ellipse. You’ve derived it from scratch!!

# Latest Mathematics Proofs: August 2016

Recently I discovered a few more proofs, some related to A Level mathematics.  You can access these proofs by clicking on the links below.

Derive the formula of an ellipsehttps://plus.google.com/+mathsvideosforweb/posts/ZF8D3ghSNRD

Discover the distance between point A and B on the edge of a circle: https://plus.google.com/+mathsvideosforweb/posts/9fmuEJ4qeXY

Find the area of a sector within a circle: https://plus.google.com/+mathsvideosforweb/posts/cHt9t9PeSfg

Formula for a torus, derived from scratch: https://plus.google.com/+mathsvideosforweb/posts/TS84TL44BYd

Create the mathematical singularity shown in movie documentaries: https://plus.google.com/+mathsvideosforweb/posts/7p2onSzNYte

Parameterised formula for a torus derived from scratch: https://plus.google.com/+mathsvideosforweb/posts/AZrVMiPVdbv

Derive the formula for an ellipsoid (normal and parameterised) from scratch: https://plus.google.com/+mathsvideosforweb/posts/Wijf94ztn3v

Other news:

If you love mathematical proofs, please feel free to join my ‘mathematics proofs’ Google community at: https://plus.google.com/communities/106007058741903558109

I’ve also got a new Facebook page related to stuff about the universe and also mathematics. You can add me as a friend by accessing this link: http://www.facebook.com/tiago.hands