Math, asked by sethuprane49381, 9 months ago

Let p be a point on the parabola, x2 = 4y. If the distance of p from the centre of the circle, x2 + y2 + 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at p, is :

Answers

Answered by Agastya0606
1

Given: P be a point on the parabola, x^2 = 4y,  centre of the circle, x^2 + y^2 + 6x + 8 = 0 is minimum.

To find: The equation of the tangent to the parabola at P

Solution:

  • Let the point P on parabola be (2t , t^2) and the centre be (-g, -f)
  • Then the centre will be: (-3, 0)
  • Now slope for line from centre to P will be:

             y2 - y1 / x2 - x1 = t^2 - 0 / 2t + 3       ...................(i)

  • So the slope of tangent to the parabola will be :

             dy/dx = x / 2 = t

             Slope of normal is: -1/t                   .......................(ii)

  • Now comparing (i) and (ii) we get:

              t^2 - 0 / 2t + 3  =   t

              t^3 + 2t + 3 = 0

              (t+1)(t^2 - t + 1) = 0

  • So the real root will be: -1
  • So the coordinates of P will be: (2t , t^2)  =  (-2 ,1 )
  • Now slope of tangent to parabola at P will be:

              dy/dx = x/2 = t = -1

  • So the equation of tangent will be:

              y-1 = -1(x+2)

Answer:

       So the equation of the tangent to the parabola at P is y + x - 3 = 0

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