A bar of length 20 units moves with its ends | vedclass

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(D) Let the two fixed lines be the coordinate axes $OX$ and $OY$. Let the bar be $AB$ with length $L = 20$.Let $P$ be a point on $AB$ such that $AP =

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$\frac{\sqrt{5}}{3}$

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(D) Let the two fixed lines be the coordinate axes $OX$ and $OY$. Let the bar be $AB$ with length $L = 20$.Let $P$ be a point on $AB$ such that $AP = 8$ and $PB = 20 - 8 = 12$.Let $\angle OAB = 	heta$. Then the coordinates of $P(x, y)$ are given by:$x = PB \cos 	heta = 12 \cos 	heta$$y = AP \sin 	heta = 8 \sin 	heta$Thus,$\cos 	heta = \frac{x}{12}$ and $\sin 	heta = \frac{y}{8}$.Using $\cos^2 	heta + \sin^2 	heta = 1$,we get:$\frac{x^2}{12^2} + \frac{y^2}{8^2} = 1$This is the equation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a = 12$ and $b = 8$.Since $a > b$,the eccentricity $e$ is given by $e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{8^2}{12^2} = 1 - \frac{64}{144} = 1 - \frac{4}{9} = \frac{5}{9}$.Therefore,$e = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.

$A$ bar of length $20$ units moves with its ends on two fixed straight lines at right angles. $A$ point $P$ marked on the bar at a distance of $8$ units from one end describes a conic whose eccentricity is

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