Ellipse and Line
      Polar and pole of the ellipse
         Equation of the polar of the given point
      Ellipse and line examples
Polar and pole of the ellipse
If from a point A(x0, y0), exterior to the ellipse, drawn are tangents, then the secant line passing through the contact points, D1(x1, y1) and D2(x2, y2) is the polar of the point A. The point A is called the pole of the
polar, as shows the right figure.   
Coordinates of the point A(x0, y0) must satisfy the            equations of tangents, thus
t1 ::   b2x0x1 + a2y0y1 = a2b2
t2 ::   b2x0x2 + a2y0y2 = a2b2
and after subtracting t1 - t2   
b2x0(x2 - x1) + a2y0(y2 - y1) = 0
obtained is
the slope of the secant line through points of contact D1and D2.
Thus, the equation of the secant line    and after rearranging
b2x0x + a2y0y = b2x0x1 + a2y0y1,  since   b2x0x1 + a2y0y1 = a2b2
follows b2x0x + a2y0y = a2b2 the equation of the polar p of the point A(x0, y0).
Ellipse and line examples
Example:  At a point A(-c, y > 0) where c denotes the focal distance, on the ellipse 16x2 + 25y2 = 1600 drawn is a tangent to the ellipse, find the area of the triangle that tangent forms by the coordinate axes.
Solution:   Rewrite the equation of the ellipse to the standard form  16x2 + 25y2 = 1600 |  1600
We calculate the ordinate of the point A by plugging the abscissa into equation of the ellipse                      
x = -6  =>    16x2 + 25y2 = 1600,
or, as we know that the point with the abscise             x = - c  has the ordinate equal to the value of the         semi-latus rectum,
The area of the triangle formed by the tangent and the coordinate axes,
Example:  Find a point on the ellipse x2 + 5y2 = 36 which is the closest, and which is the farthest from the line 6x + 5y - 25 = 0.
Solution:   The tangency points of tangents to the ellipse which are parallel with the given line are, the 
closest and the farthest points from the line.
Rewrite the equation of the ellipse to determine its axes,
Tangents and given line have the same slope, so
Using the tangency condition, determine the intercepts c,
therefore, the equations of tangents,
Solutions of the system of equations of tangents to the ellipse determine the points of contact, i.e., the 
closest and the farthest point of the ellipse from the given line, thus
Example: Determine equation of the ellipse which the line -3x + 10y = 25 touches at the point P(-3, 8/5).
Solution:   As the given line is the tangent to the ellipse, parameters, m and c of the line must satisfy the tangency condition, and the point P must satisfy the equations of the line and the ellipse, thus
Example:  The line x + 14y - 25 = 0 is the polar of the ellipse x2 + 4y2 = 25. Find coordinates of the pole.
Solution:   Intersections of the polar and the ellipse are points of contact of tangents drawn from the pole P to ellipse, thus solutions of the system of equations,
(1)  x + 14y - 25 = 0   =>   x = 25 - 14y  =>  (2)
    (2)  x2 + 4y2 = 25
   (25 - 14y)2 + 4y2 = 25
2y2 - 7y + 6 = 0,      y1 = 3/2 and  y2 = 2
y1 and y2  =>  x = 25 - 14y,   x1 = 4 and  x2 = -3.
Thus, the points of contact  D1(4, 3/2) and D2(-3, 2).
The equations of the tangents at D1 and D2,
The solutions of the system of equations t1 and t2 are the coordinates of the pole P(1, 7/2).
Example:  Find the equations of the common tangents of the curves  4x2 + 9y2 = 36 and  x2 + y2 = 5.
Solution:  The common tangents of the ellipse and the circle must satisfy the tangency conditions of these curves, thus
Pre-calculus contents H
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