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Find the locus of the midpoints of the portion of the normal to theparabola `y^2=4a x`intercepted between the curve and the axis. |
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Answer» Correct Answer - `y^(2)=a(x-a)` Normal at `P(at^(2),2at)" is "t=-tx+2at+at^(3)`. It meets the axis y=0 at `G(2a+at^(2),0)`. If (x,y) is the midpoint of PG, then `2x=2a+at^(2)+at^(2),2y=2at` Eliminating t, we have `x-a=at^(2)-a((y)/(a))^(2)` `or" "y^(2)=a(x-a)` which is the required locus. |
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