The ends of 20 cm long rod moves on two mutually perpendicular lines. If a point on this rod at 4 cm distance from end then from the eccentricity of the ellipse formed by moving that point.

Explanation

Let the length of rod AB=20cm which moves on two mutually perpendicular lines OX and OY. Let point P(x,y) be at a didtance of 4 cm from end A.
`:.` PB=AB-AP=20-4=16cm
Let PM and PN are perpendiculars from p to x-axis and y-axis respectively.
`:." "PM=yandPN-x`
Let `anglePAM=theta" "rArr" "angleBPN=theta`
in `DeltaAPM,sintheta=(y)/(4)and"in"DeltaBPN,costheta=(x)/(16)`
Now `cos^(2)theta+sin^(2)=1`
`rArr(x^(2))/(16^(2))+(y^(2))/(4^(2))=1`, which is the equation of the ellipse.
Here a=16 and b=4
`:." "b^(2)=a^(2)(1-e^(2))`
`"From "16=256(1-e^(2))`
`rArr" "1-e^(2)=(1)/(16)`
`rArr" "e^(2)=(15)/(16)`
`rArr" "e=(sqrt(15))/(4)`.