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In the following figure, find the locus of centroid of triangle PAB, where AP perpendicular to PB. |
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Answer» Solution :`:.`Slope of AP `=(2)/(t)` `rArr`Slope of BP `=-(t)/(2)` So, equation of line BP is `y-2t=-(t)/(2)(x-t^(2))`. Putting y = 0, we get POINT B as `(t^(2)+4,0)`. Now , let centroid of `Delta PAB` be (h,k). `:.""h=(t^(2)+t^(2)+4)/(3)andk=(2t)/(3)` Eliminating 't', we get `3h-4=2((3k)/(2))^(2)` `:.""3x-4=(9Y^(2))/(2)`, which is the required locus. |
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