Conic Sections
    Parabola and Line
      Polar of the parabola
      Construction of the tangent at the point on the parabola
      Construction of the tangents from a point exterior to the parabola
      Parabola and line, examples
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.  
Construction of the tangent at the point on the parabola 
As the vertex of the parabola is the midpoint of the line  
segment whose endpoints are, the projection
P1' of the  given point to the axis of the parabola and the point B    opposite it, draw the tangent through points B and P1.
Construction of the tangents from a point exterior to the parabola
Draw the circle centered at the point A outside the         parabola through the focus.                                          
The circle intersects the directrix at points, D1 and D2
Tangents from the point A are then perpendicular          
bisectors of the line segments,
D1F and D2F.             
Thus, the tangency points, P1 and P2 are equidistant    from the focus and the directrix.                                   
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, 
Intermediate algebra contents
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