A straight rod of given length slides between two fixed bars which include an ingle of`90^0` . Show that the locus of a point on the rod which divides it in a given ratio is an ellipse. If this ratio be 1/2, show that the eccentricity of the ellipse is `sqrt(2)//3.`

Explanation

Let PQ be the line.

So the lines be divided by (h,k)

Thus, H=`2acostheta/3` and `k=2asintheta/3`

`3h/(2a)=costheta` and `3k/2a=sintheta`

So,`cos^2theta+sin^2theta=1`

=`(9h^2)/(4a^2)+(9k^2)/(a^2)=1`

=`x^2/(4a^2/9)+y^2/(a^2/9)=1`

Or `e=sqrt(1-b^2/a^2)`

...