InterviewSolution
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InterviewSolution
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Perpendiculars are drawn from two points on axis of parabola `y^2=4ax` to a tangent to the parabola. The two point are situated at a distance of d from focus. if the perpendicular distances from the points are `p_1` ND `P_2` THEN FIND THE VALUE OF `P_1^2`-`P_2^2` |
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Answer» `y^2=4ax` `y(2at)=2a(x+at^2)` `ty=x+at^2` `x-ty+at^2=0-(1)` `P_1=(|(a+d)-0+at^3|)/sqrt(1+t^2)` `P_2=(|(a-d)-0+at^3|)/sqrt(1+t^2)` subtracting`P_2^2` from`P_1^2` `=(a+d+at^2)^2/(1+t^2)-(a-d+at^2)^2/(1+t^2)` `=(4ad+4adt^2)/(1+t^2)` `=4ad` `P_1^2-P_2^2=4ad`. |
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