In order to derive the equation of an ellipse centered at the origin, consider an ellipse that is elongated horizontally into a rectangular coordinate system and whose center is placed at the origin.
The foci are on the x-axis at (-c,0) and (c,0) and the vertices are also on the x-axis at (-a,0) and (a,0)
Let (x,y) be the coordinates of any point on the ellipse. According to the definition of an ellipse, d_{1} + d_{2} = constant.
It is very important to understand the meaning of the equation
d_{1} + d_{2} = constant in order to derive the equation of an ellipse.
We will use the distance formula a few times in order to find different expressions d_{1} and d_{2} .
What will be a little tricky is to find what the constant is equal to. First, we will look for the constant, so gather all of your mental energy. The meaning of the definition is that no matter where you put (x,y), you will always get the same constant.
Since it does not matter where we put (x,y), we should put it on the x-axis. Why? It is because it will be easier to find what the constant is equal to. So here we go! The figure below shows the point (x,y) on the x-axis.
Notice that d_{1} is the distance between (-c, 0) and (a, 0) and d_{2} is the distance between (c, 0) and (a, 0)
Using the distance formula,
$$ d_1 = \sqrt{(a--c)^2 +(0-0)^2} = \sqrt{(a+c)^2} = a + c $$ |
$$ d_2 = \sqrt{(a-c)^2 +(0-0)^2} = \sqrt{(a-c)^2} = a - c $$ |
d_{1} + d_{2} = c + a + a - c = 2a
We have reached a milestone trying to derive the equation of an ellipse.
Now, we need to find expressions for d_{1} and d_{2}
For this, we will need this figure.
Notice that when using this figure, d_{1} is the distance between (-c, 0) and (x, y) and d_{2} is the distance between (c, 0) and (x, y)
$$ d_1 = \sqrt{(x--c)^2 +(y-0)^2} \\ d_1 = \sqrt{(x+c)^2 + y^2} $$ |
$$ d_2 = \sqrt{(x-c)^2 +(y-0)^2} \\ d_2 = \sqrt{(x-c)^2 + y^2} $$ |
Let us put it all together! d_{1} + d_{2} = 2a
$$ \sqrt{(x+c)^2 + y^2} + \sqrt{(x-c)^2 + y^2} = 2a $$ |
$$ x^2 + 2cx + c^2 + y^2 = 4a^2 -4a \sqrt{(x-c)^2 + y^2} + x^2 - 2cx + c^2 + y^2 $$ |
$$ 2cx = 4a^2 -4a \sqrt{(x-c)^2 + y^2} - 2cx $$ |
c^{2}x^{2} - 2a^{2}cx + a^{4} = a^{2}(x^{2 }- 2cx + c^{2} + y^{2})
c^{2}x^{2} - 2a^{2}cx + a^{4} = a^{2}x^{2 }-2a^{2}cx + a^{2}c^{2} + a^{2}y^{2}
Simplify by getting rid of - 2a^{2}cx
c^{2}x^{2} + a^{4} = a^{2}x^{2 } + a^{2}c^{2} + a^{2}y^{2}
Rearrange the terms so that terms that have the variable x or the variable y are on the same side of the equation and everything else is on the other side of the equation.
c^{2}x^{2} - a^{2}x^{2} - a^{2}y^{2} = a^{2}c^{2} - a^{4}
Factor out x^{2} and a^{2}
x^{2}(c^{2} - a^{2}) - a^{2}y^{2} = a^{2}(c^{2} - a^{2})^{ }
Multiply both sides by -1
x^{2}(a^{2 }- c^{2 }) + a^{2}y^{2} = a^{2}(a^{2} - c^{2})^{ }
We just need 1 little piece of information to finish this problem off! Take a look at the following figure. We now place (x,y) on the y-axis. We also define co-vertices located on the y-axis at (0,b) and (0, -b).
$$ d_1 = \sqrt{(b-0)^2 +(0--c)^2} = \sqrt{b^2 + c^2} $$ |
$$ d_2 = \sqrt{(b-0)^2 +(0-c)^2} = \sqrt{b^2 + c^2} $$ |
d_{1} + d_{2} = 2a
$$ \sqrt{b^2 + c^2} + \sqrt{b^2 + c^2} = 2a $$ |
$$ 2 \sqrt{b^2 + c^2} = 2a $$ |
Square both sides
b^{2} + c^{2} = a^{2}
Isolate b^{2}
b^{2} = a^{2} - c^{2}
Substitute b^{2} for a^{2} - c^{2} in x^{2}(a^{2 }- c^{2 }) + a^{2}y^{2} = a^{2}(a^{2} - c^{2})^{ }
x^{2}b^{2} + a^{2}y^{2} = a^{2}b^{2}^{ }
Divide both sides by a^{2}b^{2}
$$ \frac{x^2b^2}{a^2b^2} + \frac{a^2y^2}{a^2b^2} = \frac{a^2b^2}{a^2b^2} $$ |
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$ |
This is how to derive the equation of an ellipse when the horizontal major axis is centered at the origin.
May 19, 19 09:20 AM
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May 19, 19 09:20 AM
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