Ellipse The parametric equations of the ellipse Equation of a translated ellipse
Equations 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
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.
Equations of the ellipse examples
Example:  Given is equation of the ellipse 9x2 + 25y2 = 225, find the lengths of semi-major and semi-minor axes, coordinates of the foci, the eccentricity and the length of the semi-latus rectum.
 Solution:  From the standard equation we can find the semi-axes lengths dividing the given
 equation by 225, coordinates of the foci F1(-c, 0) and F2( c, 0), since  Example:  From given quantities of an ellipse determine remaining unknown quantities and write equation of the ellipse, Solution:    a)  Using therefore, the semi-minor axis the linear eccentricity the semi latus rectum and the equation of the ellipse   the eccentricity and the equation of the ellipse  the semi latus rectum and the equation of the ellipse d) unknown quantities expressed through given values, Example:   Find the equation of the ellipse whose focus is F2(6, 0) and which passes through the point A(5Ö3, 4).
Solution:   Coordinates of the point A(5Ö3, 4) must satisfy equation of the ellipse, therefore thus, the equation of the ellipse 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:  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.   Pre-calculus contents H 