Tangent to parabola



     In this page 'Tangent to parabola' we are going to see how to get the equation of tangent for the given parabola.

A tangent to a parabola is a straight line which intersects (touches) the parabola exactly one point.We can draw only two tangents to a circle from the point outside to a parabola.

        Equation of a tangent to the parabola y²=4ax at (x₁,y₁) is

                                yy₁ = 2a(x+x)


Condition for the line y=mx+c to be a tangent to the parabola y²=4ax.

         Let (x₁,y₁) be the point where the line

                     y=mx+c-------------------(1)

touches the parabola.

        The equation of the tangent at (x₁,y₁) is

                     yy₁ = 2a(x+x₁)-------------(2)

(1) and (2) represent the same line.

                   
                     y₁/1 = 2a/m = 2ax₁/c

We get the value of (x₁, y₁) as

                   x₁ =c/m and y₁ = 2a/m.

Since (x₁, y₁) is a point on the parabola  y²=4ax,

                  we get 4a²/m² = 4ac/m

                                 c  =  a/m

is the required condition for the line y= mx+c to be a tangent to the parabola y²=4ax.

Note:

   We can draw two tangents from any point (h,k) to the parabola y²=4ax.


Example:

   Find the equation of the tangent to parabola y²= x at (4,-2).

Solution:

    y²= x is the given parabola. (4,-2) is the given point.

                           y²= x

                           y²= 4. (1/4). x      

     Hence              a = 1/4.

The equation of the tangent at (x₁,y₁) is    yy₁ = 2a(x+x₁)

          Here (x₁,y₁)     =   (4,-2) and a= 1/4.

                    y(-2)   =   2(1/4)(x+4)

                     -2y    =      x/2 +2

Multiply the whole equation by 2 we get

                    -4y    =      x+4

                    x+4y+4=0 is the required equation.

Note:

     We can find the equation of the tangent using differentiation also. By differentiating the given equation we get the slope of the tangent. Using the slop-point form we get the equation of the tangent.

     This is discussed in the differential calculus unit.


Example:

     Find the equation of the tangent to the parabola y²= 8x, which is perpendicular to the line x-3y+8=0.

Solution:

                                 y²= 8x

                                    = 4 (2)x

                  So           a  = 2.

          Slope of the given line x-3y+8=0 is

                             3y = x+8

                              y = (x+8)/3 = x/3 +8/3.

           Slope          m  = 1/3.

    The given line is perpendicular to the tangent.

    So slope of the tangent   = -1/(1/3) = -3.

We know the condition for the line y=mx+c to be the tangent to the parabola y²=4ax  is  c= a/m.

   Here  a =2 and m =-3.

        So y = (-3)x +2/(-3).

Multiplying through out by -3, -3y = 9x+2.

        That is               9x+3y+2=0 is the required tangent.


      Students and teachers can guide the students to follow the examples discussed in the page 'Tangent to parabola'. They can practice the problems in the same method. If you have any doubt you can contact us through mail, we will help you to clear your doubts.




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