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InterviewSolution
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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. |
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Answer» 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)`. |
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