If P_1 P_2 and P_3 P_4 are two focal chords of the parabola y^2=4 a x then the chords P_1 P_3 and P_2 P_4 intersect on the

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MARKS Content Team · Accepted Answer

Correct answer: (A) directrix of the parabola. Let P_i (a_i^2, 2 a t_i ), i=1,2,3,4 then, t_1 t_2=t_3 t_4=-1 equation of P_1 P_3 : (t_1+t_3 ) y=2 x+2 a t_1 t_3 ...(1) P_2 P_4: (t_2+t_4 ) y=2 x+2 a t_2 t_4 ...(2) Putting x=-a in equation (1) gives y=(2 a t_1 t_3-2 a)/(t_1+t_3) Putting x=-a in equation (2) gives y=(2 at_2 t_4-2 a)/(t_2+t_4)=((2 a)/(t_4 t_3)-2 a)/(-1/t_1-1/t_3)=(2 a (1-t_1 t_3 ))/(- (t_1+t_3 ))=(2 a (t_1 t_3-1 ))/((t_1+t_3 )) ∴ These lines meet at (-a, (2 a (t_1 t_3-1 ))/(t_1+t_3) )

If $P_1 P_2$ and $P_3 P_4$ are two focal chords of the parabola $y^2=4 a x$ then the chords $P_1 P_3$ and $P_2 P_4$ intersect on the

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