# the line joining (5,0) to (10 cosθ ,10sinθ ) is dividedinternally in the ratio 2:3 at P.If varies,then the locus of P is.

Prajwal kr
49 Points
9 years ago

I give you the idea:

Let any point on locus be (h,k)

Apply section formula. Get a relation such that you get

cosθ= some value in terms of h........................................(1)

sinθ= some value in terms of k....................................(2)

Square the equaltions and add them.

sinθ and

 cosθ are elimintated. This will be the locus.

Chakshu Shah
33 Points
3 years ago
(h,k) =[2(10$\large \cos \theta$)+3(5)/5 , 2(10$\large \sin \theta$)+3(0)/5]
=[4$\large cos \theta$+3 , 4$\large \sin \theta$]
$\large \therefore$ h = 4$\large cos \theta$+3 and also k= 4$\large \sin \theta$
So, $\large cos \theta$= h-3/4......................(1)
$\large \sin \theta$=k/4.........................(2)
Squaring and adding both the equantions,
$\large \therefore$ $\large \cos ^{2}\theta +\sin ^{2}\theta$=$\large h^{2}+9-6h +k^{2}$$\large /16$
$\large \therefore$ $\large 16=h^{2}+ k^{2}-6h+9$
$\large \therefore h^{2}+k^{2}-6h-7=0$
If $\large \theta$ varies then locus of P is a circle