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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?

Arun
25763 Points
3 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

Regards
Harsh Vardhan
13 Points
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 parabola
L^2=4ak
(3y1)^2=4ax1
9y1^2=4ax1
Hence the locus of the point of trisection is
9y^2=4ax
9 months ago
Dear student,

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

Hope it helps.
Thanks and regards,
Kushagra