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In the following figure, find the locus ...

In the following figure, find the locus of centroid of triangle PAB, where AP perpendicular to PB.

Text Solution

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`:.` 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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