Can you find locus?

Question

Q is a point on !!(2y-1)^2=8(x-3)!!. The locus of point on !!PQ!! dividing it in 2:1 ratio is? !!P!! is !!(-6,-1)!!

  • !!y^2=12x!!
  • !!y^2=4x!!
  • None

Solution:

We will give 2 approaches both of which are pretty simple and yet need us to be careful.

!!Q(x_1,y_1)!! be a point on the given parabola. Now , let us denote the point obtained after applying section formula as (h,k).
Clearly,that point divides the line PQ in the ratio 2:1 so by applying the section formula we get.
!!h = (2x_1 - 6)/3!! and !!k = (2y_1 - 1)/3!!

But we know !!(2y_1 - 1)^2 = 8(x_1 - 3)!!
Now by comparing and putting the values above in the parabola's equation, we get:
!!(3k)^2 = 8.(3h/2)!! !!k^2 = 4h/3!! That is the equation of the locus of the required point.

Alternatively a naive approach would be to find the value of !!(x_1,y_1)!!:
!!x_1 = (3h+6)/2!! and!! y_1 = (3k+1)/2!! Now putting the values of !!x_1!! and !!y_1!! in the equation of parabola, we get:
!!( 2(3k+1)/2 - 1)^2 = 8((3h+6)/2 - 3)!! !!9k^2 = 12h!! !!k^2 = 4h/3!!

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