Parabola and Line
      Common points of a line and a parabola
      Condition for a line to be the tangent to the parabola - tangency condition
      The equation of the tangent and the normal at the point on the parabola
      Parabola and line, examples
Common points of a line and a parabola
Common points of a line and a parabola we determine by solving their equations as the system of two 
equations in two unknowns,
                                       (1y = mx + c
                    (2y2 = 2px            (1)  =>   (2)      m2x2 + 2(mc - p)x + c2 = 0,
therefore, the coordinates of intersections of a line and a parabola
Condition for a line to be the tangent to the parabola - tangency condition
In the formulas for calculating coordinates of intersections there is the expression under the square root whose value determines three possible cases regarding mutual position of a line and a parabola,
so for p > 0 and,
  p - 2mc > 0  - the line intersects the parabola at two points S1(x1, y1) and S2(x2, y2),
  p - 2mc = 0  - the line is the tangent of the parabola and have one point of contact D((p - mc)/m2, p/m))
                         or by substituting  p = 2mc, the tangency point D(c/m, 2c),
  p - 2mc < 0  - the line and the parabola do not intersect.
If we write the above conditions as then 
these three cases can be explained graphically as      the relation between parameters m and c of the line    and the position of the focus F(p/2, 0).                    
At the y-intercept N of the line drawn is a                   perpendicular which intersects the x-axis at M, then  
ON = | c | and OM = | c tan a | = | m c |.
Therefore, when the point M is located;                    
- to the left of F, i.e.,  if  m c < p/2
the line intersects the parabola at S1and S2,             
  - to the right of F, i.e.,  if  m c > p/2   
the line and the parabola do not intersect,                 
            - at the focus F or if   m c = p/2                
the line is the tangent of the parabola.                       
The equation of the tangent and the normal at the point on the parabola
In the equation of the line y - y1 = m( x - x1) through the given point we express the slope m by the given 
ordinate of the tangency point,
and since the coordinate of the tangency point must        satisfy the equation of the parabola, then                       
 obtained is   y1y = p(x + x1) the equation of  
the tangent at the point P(x1, y1) on the parabola.          
Since the above equation 
can be written using coordinates of the tangency point    
   
As the slope of the normal then the equation of the normal at P(x1, y1),
  or  
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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