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Wize High School Grade 12 Pre-Calculus Textbook > Conics [BC/AB]

Ellipses

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Ellipses

An ellipse is the set of points in the plane in which the sum of distances from two fixed points F1F_1 and F2F_2 is a constant.

The foci have the coordinates (−c,0)(-c,0) and (c,0)(c,0) where c2=a2−b2c^2=a^2-b^2

The vertices have the coordinates (−a,0)(-a,0) and (a,0)(a,0) and are the endpoints of the major axis.

The major axis is the line segment joining the vertices and is the larger diameter of the ellipse.

The minor axis is the smaller diameter of the ellipse.



Let the sum of the distances from a point on the ellipse P(x,y)P(x,y) to the foci be 2a>02a>0.

Then, P(x,y)P(x,y) is a point on the ellipse when:

∣PF1∣+∣PF2∣=2a(x+c)2+y2+(x−c)2+y2=2a(x−c)2+y2=2a−(x+c)2+y2(x−c)2+y2=4a2−4a(x+c)2+y2+(x+c)2+y2x2−2xc+c2+y2=4a2−4a(x+c)2+y2+x2+2xc+c2+y20=a2−a(x+c)2+y2+xca(x+c)2+y2=a2+cxa2(x2+2cx+c2+y2)=a4+2a2cx+c2x2a2x2+2a2cx+a2c2+a2y2=a4+2a2cx+c2x2a2x2−c2x2+a2y2=a4−a2c2x2(a2−c2)+a2y2=a2(a2−c2)\begin{array}{rcl} |PF_1|+|PF_2|&=&2a\\\\ \sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}&=&2a\\\\ \sqrt{(x-c)^2+y^2}&=&2a-\sqrt{(x+c)^2+y^2}\\\\ (x-c)^2+y^2&=&4a^2-4a\sqrt{(x+c)^2+y^2}+(x+c)^2+y^2\\\\ x^2-2xc+c^2+y^2&=&4a^2-4a\sqrt{(x+c)^2+y^2}+x^2+2xc+c^2+y^2\\\\ 0&=&a^2-a\sqrt{(x+c)^2+y^2}+xc\\\\ a\sqrt{(x+c)^2+y^2}&=&a^2+cx\\\\ a^2(x^2+2cx+c^2+y^2)&=&a^4+2a^2cx+c^2x^2\\\\ a^2x^2+2a^2cx+a^2c^2+a^2y^2&=&a^4+2a^2cx+c^2x^2\\\\ a^2x^2-c^2x^2+a^2y^2&=&a^4-a^2c^2\\\\ x^2(a^2-c^2)+a^2y^2&=&a^2(a^2-c^2)\\\\ \end{array}
Notice in the above figure that 2c<2a2c<2a and so c<ac<a.

We can therefore conclude that (a2−c2)>0(a^2-c^2)>0.

Let b2=a2−c2b^2=a^2-c^2 for the sake of convenience.

Then, x2(a2−c2)+a2y2=a2(a2−c2)x^2(a^2-c^2)+a^2y^2=a^2(a^2-c^2) becomes x2b2+a2y2=a2b2x^2b^2+a^2y^2=a^2b^2 we can be simplified to:

x2a2+y2b2=1\boxed{\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1}

which is the equation of an ellipse.


Wize Tip
The transformed form for the equation of an ellipse is(x−h)2a2+(y−k)2b2=1\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1.

Example: Ellipses

Find the endpoints of the major axis & minor axis, the length of the major axis & minor axis, and the points for the foci.

x225+y216=1\dfrac{x^2}{25}+\dfrac{y^2}{16}=1

The center of the ellipse is at (0,0).(0,0).

Then,

a2=25  →  a=5b2=16  →  a=4a^2=25~~{\color{red}\rightarrow}~~a=5\newline{} b^2=16~~{\color{red}\rightarrow}~~a=4


The endpoints of the major axis are (−5,0)(-5,0) and (5,0)(5,0) and the length is 10 units.

The endpoints of the minor axis are (−4,0)(-4,0) and (4,0)(4,0) and the length is 8 units.

The endpoints for the foci are:

c2=a2−b2=25−16=9c=±3\begin{array}{rcl} c^2&=&a^2-b^2\\ &=&25-16\\ &=&9\\ c&=&\pm3 \end{array}

(−3,0)(-3,0) and (3,0)(3,0)

Practice: Ellipses

Match the equation to the correct graph.
A.
x225+y249=1\dfrac{x^2}{25}+\dfrac{y^2}{49}=1
B.
(x−1)24+(y−3)236=1\dfrac{(x-1)^2}{4}+\dfrac{(y-3)^2}{36}=1
C.
(x+2)28+y212=1\dfrac{(x+2)^2}{8}+\dfrac{y^2}{12}=1
D.
x281+(y−4)216=1\dfrac{x^2}{81}+\dfrac{(y-4)^2}{16}=1








Practice: Ellipses

Find the following information for the conic 25x2−150x+36y2+72y−639=025x^2-150x+36y^2+72y-639=0.

Practice: Ellipses

Find the equation of an ellipse with the following characteristics:
  • Foci (2,−2)(2,-2) and (4,−2)(4,-2)
  • Vertices (1,−2)(1,-2) and (5,−2)(5,-2)

Extra Practice
Conics
Sketch a graph of the function x2+8x+3y2−6y+7=0x^2+8x+3y^2-6y+7=0.
Ellipses
Practice: Ellipses
Find the equation of an ellipse with the following characteristics:
Foci (−4,3)(-4,3) and (9,3)(9,3)
Ellipses
Practice: Ellipses
Find the equation of an ellipse with the following characteristics:
Foci (5,−4)(5,-4) and (8,−4)(8,-4)
Ellipses
Practice: Ellipses
Find the following information for the conic 4x2+4x+2y2−8y−9=04x^2+4x+2y^2-8y-9=0.
Ellipses
Practice: Ellipses
Find the following information for the conic 49x2−7x+81y2+9y−450=049x^2-7x+81y^2+9y-450=0.
Ellipses
Practice: Ellipses
Match the equation to the correct graph.
Ellipses
Practice: Ellipses
Match the equation to the correct graph.