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  • Circles are drawn taking any two focal chords of the parabola as diameters. Then, the equation of com

Circles are drawn taking any two focal chords of the parabola y^2=4 a x as diameters. Then, the equation of common chord is equal to

Option: 1

a\left(t_3^2+t_4^2-t_1^2-t_2^2\right) x+a\left(t_3+t_4-t_1-t_2\right) y=0


Option: 2

a\left(t_3^2+t_4^2+t_1^2-t_2^2\right) x-2 a\left(t_3+t_4-t_1+t_2\right) y=0


Option: 3

a\left(t_3^2+t_4^2-t_1^2-t_2^2\right) x+2 a\left(t_3+t_4-t_1-t_2\right) y=0


Option: 4

None of the above 


Answers (1)

best_answer

(c) Let PQ and RS be the two focal chord of the parabola and let P, Q, R and S be the points t_1, t_2, t_3, t_4 respectively, then t_1 t_2=-1 and t_3 t_4=-1. Clearly, on PQ as diameter is

\begin{aligned} & \left(x-a t_1^2\right)\left(x-a t_2^2\right)+\left(y-2 a t_1\right)\left(y-2 a t_2\right)=0 \\ \\\Rightarrow & x^2+y^2-a\left(t_1^2+t_2^2\right) x-2 a\left(t_1+t_2\right) y-3 a^2=0\: \: \: \: \: \: \: \: \: ...(i) \end{aligned}

and circle on RS as diameter is 

x^2+y^2-a\left(t_3^2+t_4^2\right) x-2 a\left(t_3+t_4\right) y-3 a^2=0\: \: \: \: \: \: \: \: \: ...(ii)

\therefore Equation of the common chord is 

a\left(t_3^2+t_4^2-t_1^2-t_2^2\right) x+2 a\left(t_3+t_4-t_1-t_2\right) y=0

Posted by

himanshu.meshram

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