The locus of a point P which divides the line | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(b) Let the coordinates of the point $P$ which divides the line joining $(1, 0)$ and $(2\cos 	heta ,\,2\sin 	heta )$in the ratio $2:3$ be $(h,k)$. T

Vedclass · Accepted Answer

Circle

Vedclass · Answer

(b) Let the coordinates of the point $P$ which divides the line joining $(1, 0)$ and $(2\cos 	heta ,\,2\sin 	heta )$in the ratio $2:3$ be $(h,k)$. Then, $h = \frac{{4\cos 	heta + 3}}{5}$and $k = \frac{{4\sin 	heta }}{5}$

==> $\cos 	heta = \frac{{5h - 3}}{4}$and $\sin 	heta = \frac{{5k}}{4}$

==>${\left( {\frac{{5h - 3}}{4}} ight)^2} + {\left( {\frac{{5k}}{4}} ight)^2} = 1$==>${(5h - 3)^2} + (5{k^2}) = 16$

Therefore locus of $(h,k)$ is ${(5x - 3)^2} + {(5y)^2} = 16$,which is $a$ circle.

The locus of a point $P$ which divides the line joining $(1, 0)$ and $(2\cos \theta ,2\sin \theta )$ internally in the ratio $2 : 3$ for all $\theta $, is a

Similar Questions

Let $A B C D$ be a square of side length $1$ . Let $P, Q, R, S$ be points in the interiors of the sides $A D, B C, A B, C D$ respectively, such that $P Q$ and $R S$ intersect at right angles. If $P Q=\frac{3 \sqrt{3}}{4}$, then $R S$ equals

The equation of the line which makes right angled triangle with axes whose area is $6$ sq. units and whose hypotenuse is of $5$ units, is

If the middle points of the sides $BC,\, CA$ and $AB$ of the triangle $ABC$ be $(1, 3), \,(5, 7)$ and $(-5, 7)$, then the equation of the side $AB$ is

The ends of the base of an isosceles triangle are at $(2a,\;0)$ and $(0,\;a).$ The equation of one side is $x=2a$ The equation of the other side is

The area of the triangle formed by the line $x\sin \alpha + y\cos \alpha = \sin 2\alpha $and the coordinates axes is