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`        If PQ is the double ordinate of the parabola y2=4ax then what is the locus of its point of trisection?`
2 years ago

```							Dear Vijaya The equation y^2 = 4ax can be parametrised as: x = at^2, y = 2at. If the extremities of the double ordinate are (at^2, 2at) and (at^2, - 2at), then its points of trisection have ordinates y1 and y2 where: y1 = - 2at + 4at / 3 = 2at / 3 y2 = - 2at + 8at / 3 = - 2at / 3 For both of these points: y^2 = 4a^2 t^2 / 9 y^2 = (4a / 9)(at^2) y^2 = 4(a / 9)x. The locus is the parabola y^2 = 4bx where: b = a / 9.hence locus -y^2 = (4/9) ax RegardsArun(askIITians forum expert)
```
2 years ago
```							Let the extremities of double ordinate be (k,l) and (k,-l)Let P(x1,y1) be any point on the double ordinate which divides the line in the ratio 1:2 P(x1,y1)=1(k)+2(k)/1+2,1(-1)+2(l)/1+2=(K,l/3)Here x1=k y1=l/3,l=3y1(K,l)=(x1,3y1)As (k,l) lies on the parabolaL^2=4ak(3y1)^2=4ax19y1^2=4ax1Hence the locus of the point of trisection is 9y^2=4ax
```
one year ago
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