A rod AB of length 3 rests on a wall as follows:
P is a point on AB such that AP:PB=1:2. If the rod slides along the wall, then the locus of the point lies in $4x^2+y^2=4$- Prove it.
My attempt
Although I could only guess that the locus will be an ellipse by intuition and figure but no mathematical basis to prove my point. Please please help. I am clueless here.
On
The question becomes quite easy while assuming the polar format for solving the question and finding the locus. The given point "P", has an x co-ordinate of l/3cos(k), where k is the angle made by the rod with the horizontal, and length of rod is taken l, similarly the y co-ordinate of the rod is 2l/3sin(k), as now 3x/l is cos(k), and 3y/2l is sin(k); now squaring and adding for achieving the pythagorean identity, we get 9x^2/l^2 + 9y^2/4l^2 =1, as l=3, hence the locus decomposes to 4x^2+y^2=4.
sorry for the sluggish explanation I am a beginner and unaware of using Math Jax.
Choose the origin as the intersection of the two walls.
Let $A=(0,a)$ and $B=(b,0)$ be the coordinates of the end points of the ladder in a general configuration.
Then, the point $P$ is $(\frac{b}{3}, \frac{2a}{3})$. (Why?)
As the ladder slips around, $a$ and $b$ can change but since the length is $3$, we must have $$ a^2+b^2=9$$ (Why?)
Substitute $x=\frac{b}{3}$ and $y=\frac{2a}{3}$ into the above constraint to get your required locus.