1.

The power of the origin w.r.t the circle on a focal chord of `y^2 = 4ax` as a diameter is

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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