Derivations
of Equations
Ellipse
We'll use the
following two diagrams for an ellipse in the derivation:
Recall the definition
of an ellipse requires that d1
+ d2
= 2a.
Let (x,y) be the
coordinates for the point P. In other words, P can be any point on
the ellipse as shown below.
Here we wave two right
triangles. We will write the distances d1 and d2
in terms of the legs of these right triangles using Pythagorean's
Theorem then simplify the result.
Before we begin,
please note that following a mathematical derivation can be tedious
at times primarily due to the amount of Algebra involved. Just take
each step one at a time and go slowly through the derivation. I've
found that writing each step down can help you focus on the
derivation (doing this when learning proofs is one of the best
techniques I've found to learn proofs.) In many texts these Algebra
steps are left out as an exercise to the reader. Here I will show and
explain each step.
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Start with the definition of the
ellipse:
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Split radicals across the equal
sign to simplify their removal by squaring both sides of the
equation.
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Expand the binomials (c+x)2
and (c-x)2
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Eliminate the like terms shown
in red from both sides of the equation.
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Factor out the 4 from the terms
on the right hand side of the equation.
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Divide the equation by that 4.
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Square both sides of the
equation one more time to eliminate the radical.
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Expand the binomial (c-x)2
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Eliminate the parenthesis by
multiplying through with a2.
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Eliminate the like terms shown
in red on both sides of the equation.
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Group the variables and
group the constants, variables on the left and constants on the
right of the equal sign.
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Factor out the x2
and the a2 terms.
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Divide both sides by a2(a2
- c2).
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Cancel common factors.
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Substitute.
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Parabola
The definition of a parabola is:
all points whose perpendicular distances from them to the directrix
equals the distances from them to the focus.
We'll use these properties from
the definition to derive the equation of a parabola.
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The directrix is shown in blue. The vertical black line from
the directrix to P is perpendicular to the directrix.
For d1
use the absolute value here in
general, the sum could be negative. (With this parabola inverted,
y is negative, so y + c could be negative
for example, if |y| > |c|.
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The distance to point P(x,y)
from the directrix is the length of the heavy black line. This
length has two parts, |c| and |y|. In terms of the
coordinates we have |y - (-c)| = |y + c|.
So we get d1 = |y+ c|.
Now we need an expression for d2.
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Note the right triangle.
Its horizontal leg is at y =
c.
The vertical distance to
P(x,y) is just y.
Subtract these to get the
length of the vertical leg of this triangle.
Now apply Pythagorean's
Theorem
to find d2.
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To finish we equate these and
simplify the equation.
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Squaring the absolute value of
a number is the same as squaring its negative, so we can remove
the absolute value signs.
ex: | -10 |2 =
102 = 100
( -10 )2 =
(-10)(-10) = 100
Note
the common terms in red and green on both sides of the equation.
Cancel
these terms and get all terms involving y onto one side of the
equation and all terms involving x onto the other side of the
equation (by adding 2yc to both sides)
Switch
sides to get the equation for a parabola with vertex at the
origin.
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the general
equation:
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As in the ellipse a
translation of the origin from (0,0) to (h,k) results in this
equation.
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Hyperbola
The definition of a hyperbola
is: all points P
such that the difference between
the distances from P
to two other distinct points f1
and f2
is constant.
Also, this constant must be less
than 1/2 the distance between f1
and f2
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Let
this constant be 'a' and
let the notation d(i,j) mean the
distance between points i and j,
Then
we have a < 1/2 d(f1,
f2)
So
we must have 2a < d(f1,
f2)
Taken
altogether we have for the hyperbola the definition
| d(f1,
P) - d(P, f2)|
= 2a
We'll
use this definition to derive the equation for a hyperbola.
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This diagram shows the
relationships mentioned above.
You can see that the distance
from the foci f1 and f2
is larger than twice a.
d1 is
the distance from f1 to
P and d2
is the distance from P
to f2.
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This difference relationship
can be understood visually as follows:
1) follow the dark blue line
from f1 to P
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now subtract d2 from
d1 by
rotating the green segment about P so
it lies on the dark blue line and move back along the dark blue
line the length of d2.
3)
What remains of the dark blue line is shown with the dashed light
blue line.
5)
Now slide that dashed blue line onto the X axis and it will extend
from vertex to vertex with length equal to 2a.
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Recall that the distance from the origin to any focus is c.
So the distance of the horizontal leg of the blue right triangle
is |c + x| = |x + c|. (Absolute value since x could be
negative which could make x + c negative.)
The vertical leg of the right triangle is just y.
Apply Pythagorean's Theorem.
Recall that the square of the absolute value of a number is
just the square of that number.
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The horizontal leg of this green right triangle has length: the
x coordinate of P, minus c.
As before this difference could be
negative so we must use |x - c|.
The vertical leg is just y.
Use Pythagorean's Theorem.
Recall that the square of the
absolute value of a number is just the square of that number.
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Now we're ready to derive the equation of a hyperbola.
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Start
with the definition of a hyperbola.
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Two
possibilities exist for the absolute value of d1 - d2
:
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This
means – (d1 - d2) = 2a or (d1
- d2) = 2a.
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We
can use the symbol ± to cover both cases.
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Substitute
the expressions for d1 and d2.
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Isolate
one radical to make the squaring operation easier, then square
both sides of the equation.
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We
are left with two binomials to expand and one radical to
eliminate.
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Expand the
binomials and note the common terms.
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Eliminate
the common terms shown in red.
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Combine
the like term 2xc.
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Isolate the
radical so that squaring both sides of the equation eliminates it.
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4
is common to all three terms, divide the equation by 4 to remove
it.
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Now
square both sides of the equation.
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Expand
the binomial (x - c)2.
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Clear
the grouping symbols on the right hand side of the equation by
multiplying through the expression by a2.
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Eliminate
the common term -2xca2.
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Notice
the pattern c2 - a2. Factoring x2
out of x2c2 - a2x2 and
a2
out of a2c2 - a4 gives:
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Now
divide the equation by a2(c2 - a2).
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Cancel
common factors in each fraction.
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Recall
that a, b, and c are related by the equation:
c2
- a2 = b2
so
we may substitute b2 for the c2 - a2
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