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JEE MainMathematicsParabola

A curve C passes through the point (2,4) . The tangent to the curve at any point P(x,y) meets the y -axis at R . If the y -coordinate of R is equal to the negative of the square of the x -coordinate of P , then the shortest distance between the curve C and the line x - y - 2 = 0 is

Options

  1. A7 4 2
  2. B2
  3. C2 2
  4. D7 2 2

Correct answer

A. 7 4 2

Step-by-step solution

Let the point on the curve be P(x,y) . The equation of the tangent at P is Y - y = dy dx (X - x) . To find the y -intercept R , put X = 0 : Y_R = y - x dy dx . Given that Y_R = -x^2 , we have: y - x dy dx = -x^2 dy dx - 1 x y = x . This is a linear differential equation of the form dy dx + Py = Q , where P = - 1 x and Q = x . The integrating factor is IF = e^ (- 1 x ) dx = e^ - x = 1 x . The solution is given by: y 1 x = x 1 x , dx = 1 , dx = x + C y = x^2 + Cx . Since the curve passes through (2,4) : 4 = 2^2 + C(2

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