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.

Useful trigonometric formulas for finding areas of circles

If you’re trying to find the area of a circle using integration methods, then these trigonometric formulas are going to be very useful:

First formulas:

$\sin ^{ 2 }{ \theta +\cos ^{ 2 }{ \theta =1 } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta +{ r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 } } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta ={ r }^{ 2 }-{ r }^{ 2 }\cos ^{ 2 }{ \theta } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta ={ r }^{ 2 }\left( 1-\cos ^{ 2 }{ \theta } \right) }$

Second formulas:

$\sin ^{ 2 }{ \theta +\cos ^{ 2 }{ \theta =1 } } \\ \\ \therefore \quad { r }^{ 2 }\sin ^{ 2 }{ \theta +{ r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 } } } \\ \\ \therefore \quad { r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 }-{ r }^{ 2 }\sin ^{ 2 }{ \theta } } \\ \\ \therefore \quad { r }^{ 2 }\cos ^{ 2 }{ \theta ={ r }^{ 2 }\left( 1-\sin ^{ 2 }{ \theta } \right) }$

These formulas are to be used when you have to transform the expression:

${ y }=\sqrt { { r }^{ 2 }-{ x }^{ 2 } }$

You can either make:

$x=r\sin { \theta }$

Or…

$x=r\cos { \theta }$

The choice is yours. 🙂

If a/b=c/d, then a/b=(a+c)/(b+d) mathematical proof

In this post I’ll be proving to you that if:

a/b=c/d

Then:

a/b=(a+c)/(b+d)

To prove this, the first thing you need to know is that:

$\frac { a }{ b } =\frac { c }{ d } \\ \\ \therefore \quad ad=bc$

The rest is just mathematical / algebraic trickery… Let me show you…

Proof 1:

$LHS\\ \\ =\frac { a }{ b } \\ \\ =\frac { a }{ b } \cdot 1\\ \\ =\frac { a }{ b } \cdot \frac { \left( b+d \right) }{ \left( b+d \right) } \\ \\ =\frac { ab+ad }{ b\left( b+d \right) } \\ \\ =\frac { ab }{ b\left( b+d \right) } +\frac { ad }{ b\left( b+d \right) } \\ \\ =\frac { a }{ b+d } +\frac { bc }{ b\left( b+d \right) } \\ \\ =\frac { a }{ b+d } +\frac { c }{ b+d } \\ \\ =\frac { a+c }{ b+d } \\ \\ =RHS$

Proof 2:

$LHS\\ \\ =\frac { a }{ b } \\ \\ =\frac { a }{ b } \cdot 1\\ \\ =\frac { a }{ b } \cdot \frac { \left( b+d \right) }{ \left( b+d \right) } \\ \\ =\frac { ab+ad }{ b\left( b+d \right) } \\ \\ =\frac { ab+bc }{ b\left( b+d \right) } \\ \\ =\frac { b\left( a+c \right) }{ b\left( b+d \right) } \\ \\ =\frac { a+c }{ b+d } \\ \\ =RHS$

Proof 1 Video:

Proof 2 Video: