Problems in parabola



                In this page 'Problems in parabola' we are going to see some problems about tangents, normal and chord of contact to the parabola.

Problem 1:

        Find the equation of the tangent to the parabola y² = 8x, which is perpendicular to the line x-2y+6=0.

Solution:

                            y² = 8x

                                = 4(2)x

                    So here a = 2.

                 Slope of the given line x-2y+6 = 0 is

                                       2y = x+6

                                         y = x/2 +3

                  So slope of the given line is 1/2.

                  Since the tangent is perpendicular to the line, the slope of the tangent is   -2.

                  Equation of the tangent whose slope is 'm'

                                        y = mx+ a/m

                                 Here m= -2 and a =2

                  Equation of tangent is

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

                                    2x+y+1=0 is the required equation.


Problem 2:

           Find the equations of normals to the parabola y² =4ax at the end of the latus rectum.

Solution;

           Latus rectum is the line drawn through the focus. So we have x=a.

                    y²= 4a(x)

                   y² =  4a²

                    y = ± 2a

          So the latus rectum meets the parabola at (a,2a) and (a,-2a).

           Equation of normal at (x₁,y₁) to the parabola is

                             y-y₁ = -y₁/2a (x-x₁)

          So here, the equation of normal at (a,2a) is

                            y-2a = (-2a/2a) (x-a)

                              x+y-3a=0 is the  equation of normal at (a,2a).

         Equation of normal at (a,-2a) is

                            y-(-2a) = [-(-2a)/2a] (x-a)

                              y+2a  =  x-a

                              x-y-3a = 0. is the equation of normal (a,-2a).


Problem 3:

           Show that the chord of contact of tangents drawn from any point on the directrix passes through the focus of the parabola.

Solution:

           Let us take the equation of the parabola as y ² =4ax. So the equation of the directrix is x+a =0. The focus of the parabola is (a,0).

          Any point on the directrix will be (-a,k).

          We know the chord of contact equation is yy₁ = 2a(x+x₁)

          At (-a,k) the chord of contact is              yk  = 2a(x-a)

           The focus (a,0) satisfies this equation.

                           [0k = 2a(a-a) = 0].

           This proves the chord of contact yk  = 2a(x-a) passes through the focus.


Problem 4:

          Find the locus of the point whose chords of contact of tangents to the parabola y² =8x subtend a right angle at the vertex.

Solution:

          Let the moving point be P(x₁,y₁).

          Then the chord of contact is at P is yy₁ = 4(x+x₁)

          Homogenizing the equation of the parabola with the above equation

                         y²   =  8x [(yy₁-4x)/4x₁]

                     4x₁y²   =   8y₁xy-32x²

                     32x² -8y₁xy +4x₁y² =0

         This equation represents the pair of lines joining the vertex to the intersection of chord of contact and the parabola.

         Angle between these lines is given to be 90⁰.

             Coefficient of x² + coefficient of y² =0

                           32 + 4x₁ = 0

        Locus of P is 32 + 4x = 0

                                 x  = -8 is the required locus.


             Parents and teachers can guide the students to follow the problems in this page 'Problems in parabola' and master in parabola problems. If you have any doubts you can contact us through mail, we will help you to clear your doubts.



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