Conic Sections
    Ellipse
      The parametric equations of the ellipse
      Equation of a translated ellipse
         The equation of the ellipse examples
The parametric equations of the ellipse
Equation of the ellipse in the explicit form can help us to explain another construction 
of the ellipse. So, in the coordinate system draw two concentric circles of radii equal to lengths of the semi axes a and b, with the center at the origin as shows the figure.
  An arbitrary chosen line through the origin intersects the 
circle of the radius
a at the point R and the circle of radius b at M
  Then, the parallel line with the major axis through M intersects the parallel line with the minor axis through R, at a point P(x, y) of the ellipse. Proof,
in the figure, OS = x, PS = y and  
as the triangles OMN and ORS are similar, then
OM : OR  = MN : RS  or  b : a = PS : RS,
so that
It proves that the point P(x, y) obtained by the construction lies on the ellipse. This way, using the figure, we also derive
the parametric equations of the ellipse where the parameter t is an angle 0 < t < 2p
By dividing the first parametric equation by a and the second by b, then square and add them, obtained is standard equation of the ellipse.
Equation of a translated ellipse -the ellipse with the center at (x0, y0) and the major axis parallel to the x-axis.
The equation of an ellipse that is translated from its standard position can be obtained by replacing x by x0
 and y by y0 in its standard equation,  
The above equation can be rewritten into  Ax2 + By2 + Cx + Dy + E = 0.
Every equation of that form represents an ellipse if A not equal B and A · B > 0 that is, if the square terms have unequal coefficients, but the same signs.
The equation of the ellipse examples
Example:   Write equation of the ellipse passing through points A(-4, 2) and B(8, 1).
Solution:   Given points must satisfy equation of the ellipse, so
Therefore, the equation of the ellipse or   x2 + 16y2 = 80.
Example:  In the ellipse 4x2 + 9y2 = 144 inscribed is a rectangle whose vertices lie on the ellipse and whose sides are parallel with ellipse axes. Longer side, which is parallel to the major axis, relates to the shorter side as 3 : 2. Find the area of the rectangle.
Solution:  It follows from the given condition that the coordinates of vertices of the rectangle must satisfy the 
same ratio, i.e.,      x : y = 3 : 2   =>   x = 3y/2.
To determine points of the ellipse of which coordinates are in this ratio, put these variable coordinates into equation of the ellipse, 
P(3y/2, y)  =>   4x2 + 9y2 = 144
                4(3y/2)2 + 9y2 = 144  =>    18y2 = 144,
y1,2 = ±Ö144/18 = ±2Ö2,    x = 3y/2  =>  x1,2 = ±3Ö2.
Therefore, the vertices of the rectangle,
A(3Ö2, 2Ö2)B(-3Ö2, 2Ö2)C(-3Ö2, -2Ö2 and  D(3Ö2, -2Ö2).
The area of the rectangle A = 4 · x  · y = 4 · (3Ö2 ) · (2Ö2 ) = 48 square units.
Example:  Given is equation of the ellipse 4x2 + 9y2 + 24x -18y + 9 = 0,  find its center S(x0, y0), the semi-axes and intersections of the ellipse with the coordinate axes.
Solution:  Coordinates of the center and the semi-axes are shown in the equation of the translated ellipse, 
Rewrite the given equation to that form,
4(x2 + 6x) + 9(y2 - 2y) + 9 = 0
      4[(x + 3)2 - 9] + 9[(y -1)2 -1] + 9 = 0
                             4(x + 3)2 + 9(y -1)2 = 36  or
                              
therefore,  S(-3, 1)a = 3 and b = 2.
Intersections of the ellipse and the x-axis we obtain by setting  y = 0 into the equation of the ellipse, thus
4x2 + 24x + 9 = 0,    x1,2 = -3 ± 3Ö3/2,
and intersections of the ellipse with the y-axis by setting  x = 0,   =>    9y2 -18y + 9 = 0,    y1,2 = 1.
Intermediate algebra contents
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