Manipal MET2014MathematicsParabola
A B is a chord of the parabola y^2=4 a x with vertex A, B C is drawn perpendicular to A B meeting the axis at C . The projection of B C on the axis of the parabola is
Options
- Aa
- B2 a
- C4 a
- D8 a
Correct answer
C. 4 a
Step-by-step solution
Given equation of parabola is y^2=4 a x . Let the co-ordinates of B are ( a t^2, 2 a t ), then the slope of A B= 2 t Since, B C A B Slope of B C= -t 2 The equation of B C is y-2=- t 2 (x-a t^2 ) This line meets the X -axis at point O So, put y=0 , we get x=4 a+a t^2 So, the distance C D=4 a+a t^2-a t^2=4 a