Parabola and Line
      Properties of the parabola
      Polar of the parabola
      Parabola and line, examples
Properties of the parabola
  Using equations of the tangent and normal expressed by coordinates of the tangency points and the figure above;
   a)  y-intercept ct of the tangent equals half of the ordinates of the tangency point, ct = y1/2.
   b) the projection AB of the segment BP1 of the tangent to the x-axis, i.e., to the axis of the parabola, is 
       equal to twice the abscissa of the tangency point, so
                         AB = St = 2x- the line segment AB is called the subtangent.
   c) the projection AC of the segment CP1 of the normal to the x-axis is equal to the parameter p, i.e.,
                         AC = Sn =   - the line segment AC is called the subnormal.
As points B and C are x-intercepts of the tangent and the normal their abscissas we determine by solving 
corresponding equations for
y = 0, so
put y = 0 into equation of the tangent,
put y = 0 into equation of the normal,
Thus, the focus F(p/2, 0) bisects the line segment BC whose endpoints are x-intercepts of the tangent and the normal, as shows the figure above.
  The tangent at any point on the parabola bisects the angle j between focal distance and the perpendicular to the directrix and is equally inclined to the focal distance and the axis of the parabola. 
The normal at the tangency point bisect the                suplementary angle of the angle j.                           
Since,   DP1 = FP1 = r = x1 + p/2,
and       BF = FC = x1 + p/2 = r, and  DP1 || BC   
then, following triangles are congruent,                      
DBFD @ DFCP1 @ DFP1D
so, the quadrangle BFP1D is the rhombus and its     
diagonal
BP1 bisects the angle j.                           
  This property is known as the reflective property of the parabola.
A light rays coming in parallel to the axis of a             parabolic mirror (telescope), are reflected so that they all pass  through the focus. Similarly, rays originating at the  focus (headlight) will be reflected parallel to the axis.
  Tangents drawn at the endpoints of a focal chord intersect at right angles on the directrix.
a)  Solving the system of equations of tangents, 
x is the abscissa of the intersection S of tangents.
The slope of the focal chord line through tangency points,
therefore,  x = - p/2 is the abscissa of the intersection S(- p/2, y) and the equation of the directrix d.
b)  The tangent to the parabola which passes through intersection S, which lies on the directrix, must satisfy tangency condition,.e.,
Thus, satisfied is condition for perpendicularity,   m1 m2 = - 1    =>    j = 90.
Polar of the parabola
The polar p of a point A(x0, y0), exterior to the parabola y2 = 2px, is the secant through the contact points of the tangents drawn from the point A to the parabola.
The tangency points D1(x1, y1) and D2(x2, y2)  and the point A satisfy the equations of tangents,                      
t1 ::   y1y0 = p(x0 + x1)  and  t2 ::  y2y0 = p(x0 + x2).
Subtracting t2 - t1,
y0(y2  - y1) = p[(x0 + x2) - (x0 + x1)]
obtained is the slope of the polar. By plugging the slope into equation of the line through the given point               
or   y0y = y1y0 + px - px1
since  y1y0 = p(x0 + x1),
then y0y = p(x + x0) the equation of the polar.  
Parabola and line, examples
Example:  Find the angle between tangents drawn at intersection points of a line and the parabola y2 = 2px if the line passes through the focus F(7/4, 0) and its slope  m = 4/3.
Solution:   As  F(p/2, 0)  then,  p/2 = 7/4 and the equation of the parabola  y2 = 7x.
By plugging m = 4/3 and F(7/4, 0) into the equation of the line y - y1 = m(x - x1) obtained is
Intersections of the line and the parabola,                  
Substituting coordinates of S1 and S2 in we get the equation of tangents,
Slopes of tangents satisfy the perpendicularity condition,
Example:  Find the point on the parabola y2 = 9x closest to the line 9x + 4y + 24 = 0.
Solution:  The tangency point of the tangent parallel to the given line is the closest point.
9x + 4y + 24 = 0   =>  y = -(9/4)x - 6mt = - 9/4
The slope of the tangent must satisfy tangency            condition of the parabola,                                           
p = 2mc  <=   mt = - 9/4,   p = 9/2
9/2 = 2 (-9/4) c   =>    c = - 1
therefore, the tangent   t ::    y = -(9/4)x - 1.            
The solution to the system of equations of the tangent  and the parabola gives the tangency point, that is         
Example:  Given is the polar  4x + y + 12 = 0 of the parabola  y2 = -4x, find coordinates of the pole and write equations of the corresponding tangents.
Solution:  Intersections of the polar and the parabola are the tangency points of tangents drawn from the pole P. Thus, by solving the system of equations of the polar and the parabola we get the tangency points.
(1) 4x + y + 12 = 0, (1) =>  (2) (-4x - 12)2 = -4x
(2)  y2 = -4x                        4x2 + 25x + 36 = 0,
                               
Equation of the tangent at the point on the parabola,
The intersection of tangents is the pole P. Therefore, we solve the system formed by their equations, 
Pre-calculus contents H
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