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If two normals to a parabola `y^2 = 4ax` intersect at right angles then the chord joining their feet pass through a fixed point whose co-ordinates are: |
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Answer» equation of parabola `y^2=4ax` co-ordinates of P`(at_1^2,2at_1)` co-ordinates of q`(at_2^2,2at_2)` slope of normal=-t slope at `M_(N1)=-t_1` slope at `M_(N2)=-t_2` we know N1 and N2 are perpendicular product of there slope will be `(-t_1)(-t_2)=-1` `t_1*t_2=-1` from diagram we can see that P=`(at_1,2at_2)` Q=`(a/t_1^2,(2a)/t_1)` `m_(pr)=m_(pq)` `(2at_1-0)/(at_1^2-alpha)=(2at_1-2at_2)/(at_1^2-at_2^2)` `2at_1t_2+2at_12=2at_1^2-2alpha` `-alpha=at_1t_2-2alpha` `alpha=a` poiunts of R(a,o) |
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