The length of the chord of the parabola x^2=4y havingequation x- 2y+4 2=0 is:

Question

Accepted Answer

Correct answer: D. (d) JEE Main 2019 Let intersection points be P (x_1 ,y_1 ) and Q (x_2 ,y_2 ) The given equations x^2=4y … (i ) x- 2y+4 2=0 … ( ii ) Use eqn (1 ) in eqn (2 ) x- 2 x^24+4 2=0 2x^2-4x-16 2=0 x_1+x_2=2 2 ,x_1x_2=-16 ,(x_1-x_2)^2=8+64=72 Since, points P and Q both satisfy the equations (2 ) , then x_1- 2y_1+4 2=0 x_1- 2y_2+4 2=0 (x_2-x_1)= 2(y_2-y_1) ⇒(x_2-x_1)^2=2(y_2-y_1)^2 ⇒ PQ= (x_2-x_1)^2+(y_2-y_1)^2 = (x_2-x_1)^2+ (x_2-y_1)^22 =(x_2-x_1) 3 2=6 2 3 2=6 3 Hence, length of chord =6 3

$The length of the chord of the parabola x^{2} = 4 y havingequation x - 2 y + 42 = 0 is:$

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