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The power of the origin w.r.t the circle on a focal chord of `y^2 = 4ax` as a diameter is |
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Answer» PQ is a diameter. `((at_1^2+at_2^2)/2,2a(t_1+t_2)/2)` `t_1t_2=-1` `(x-a)^2+y^2=4a^2` Power=`-a(3a)=-3a^2`. |
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