JEE MainMathematicsParabola
A curve C in the xy -plane passes through the point (1,2) . The normal to the curve at any point P(x,y) meets the x -axis at G . If the x -coordinate of G always exceeds the x -coordinate of P by 2 , then the equation of the directrix of the curve C is
Options
- Ax - 1 = 0
- B2x + 3 = 0
- Cx + 2 = 0
- Dx + 1 = 0
Correct answer
D. x + 1 = 0
Step-by-step solution
Let the point on the curve be P(x,y) . The equation of the normal at P is Y - y = - dx dy (X - x) . To find the x -intercept G , put Y = 0 : -y = - dx dy (X_G - x) X_G = x + y dy dx . Given that the x -coordinate of G exceeds the x -coordinate of P by 2 , we have: X_G = x + 2 x + y dy dx = x + 2 y dy dx = 2 . Separating variables and integrating: y , dy = 2 , dx y^2 2 = 2x + C y^2 = 4x + 2C . Since the curve passes through (1,2) : 2^2 = 4(1) + 2C 4 = 4 + 2C C = 0 . The equation of the curve is y^2 = 4x , which is a