# 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!!

# Derive the formula to find areas underneath curves

In this post I’ll be revealing how you can derive the formula which can be used to find areas underneath curves, from absolute scratch. Now, just below, what you will find is the diagram that will help us produce this formula…

In this diagram what you will discover is that:

• A length a exists, which starts at the origin O and ends at a;
• A length x exists, which starts at the origin O and ends at x;
• A length x+𝛿x exists, which starts at the origin O and ends at x+𝛿x;
• A length 𝛿x exists, which starts at x and ends at x+𝛿x;
• A height y exists, which starts at the origin O and ends at y;
• A height y+𝛿y exists, which starts at the origin O and ends at y+𝛿y;
• A height 𝛿y exists, which starts at y and ends at y+𝛿y;
• There is a curve called y=f(x);
• There is an area underneath the curve called A which commences at a and ends at x;
• There is an area underneath the curve called 𝛿A which commences at x and ends at x+𝛿(Note: If you extend the distance from a to x what you get is a larger area, and the change in area can be measured. This change or difference is called 𝛿A);
• There is a rectangle that exists called QRUT. It has an area which is y𝛿x;
• There is a rectangle that exists called PRUS. It has an area which is (y+𝛿y)𝛿x;
• 𝛿A has an area larger than that of the rectangle QRUT, but smaller than that of the rectangle PRUS.

Producing the formula with the information we’ve discovered…

Ok, so we want to produce the formula which will help us find areas underneath curves from absolute scratch. At our disposal we have a helpful diagram (which we’ve looked at and analysed carefully) and we’ve been able to discover a few facts about it. I think we can now get to work…

Let’s start off by saying that:

Area QRUT < 𝛿A < Area PRUS

Which is something we already discovered.

If this is the case, we can say that:

y𝛿x < 𝛿A < (y+𝛿y)𝛿x

Now, check out what happens when we divide all the elements of this expression by 𝛿x:

What we end up with is…

Alright, now you may be saying to yourself, why do I need to know this? Well, it turns out that:

This is because as 𝛿x approaches 0, 𝛿y approaches 0 leaving (𝛿A)/(𝛿x) sandwiched between y and y+0.000000000000000001 which is virtually y.

And, also…

As a consequence, this ultimately means that:

$y=\frac { dA }{ dx }$

This is incredibly significant, because if we then integrate both sides of this equation, we get:

$\int { ydx=\int { \frac { dA }{ dx } } } dx\quad \Rightarrow \quad A=\int { ydx }$

And…

$A=\int { ydx } =F\left( x \right) +C$

Now, this equation can actually be used to find the area A underneath the curve from a to x. What we’re basically saying is that this area is equal to some function of x plus a constant. This ‘some function of x’ occurs when we integrate y which is a function of x.

Finalising the formula…

Alright so we’ve managed to latch on to something incredibly significant… We’ve got an important equation:

$A=\int { ydx } =F\left( x \right) +C$

However, it is not complete. We need to know what the constant C is. So…

If we say that at x=a the area A underneath the curve is 0, watch what happens… Look at what we get…

$O=F\left( a \right) +C$

Which means that:

$C=-F\left( a \right)$

Hence, we can conclude that:

$A=\int { ydx=F\left( x \right) } -F\left( a \right)$

And this formula can be transformed into something more fancy if we are measuring an area underneath a curve from x=a to x=b

This is probably the formula you’re most familiar with…

$A=\int _{ a }^{ b }{ ydx=F\left( b \right) } -F\left( a \right) ={ \left[ F\left( x \right) \right] }_{ a }^{ b }$

Which is the formula which can be used to find areas underneath curves.

If you are still confused and would like to go through this proof once again, please watch my video below…

Related:

Trapezium Rule Formula – Derivation

# Mathematical Art Work, including Isometric Drawings (Visualising Maths)

Over the past couple of months I’ve been drawing mathematical figures and logos. Quite recently I discovered that art can help me understand complex mathematics a little better, especially three dimensional problems. It turns out that infinitely many complex sketches can emerge out of isometric fields which can be created using pencils, rulers and compasses. Isometric fields are used by engineers and architects who model three dimensional structures. Isometric fields allow you to represent cubes and various different 3 dimensional shapes on 2 dimensional surfaces in quite spectacular fashion. Here are some pieces of work I produced using such fields…

Follow Mathematics Videos & Proofs’s board Mathematical Art & Beauty on Pinterest.

If you’d like to see more work that I’ve been able to produce, please visit my mathematics Pinterest page. I’ve thoroughly enjoyed drawing optical illusions and logos using isometric paper, and I could indeed start designing isometric logos and pieces of text to make some extra money whilst studying mathematics. It’s not that often you bump into sound business ideas, especially business ideas related to both mathematics and art.

If you are a mathematics student but have never used isometric paper before, I’d recommend downloading isometric paper via the links below and producing mathematical sketches. Isometric paper could potentially be very useful to those interested in learning more about vector geometry and extra dimensions.

And by the way, thanks for stopping by. Enjoy your new year celebrations! 🙂