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(A) The path traced by the man is an ellipse because the sum of the distances from two fixed points (foci) is constant.Let the constant sum be $2a = 1

Vedclass · Accepted Answer

$15$

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(A) The path traced by the man is an ellipse because the sum of the distances from two fixed points (foci) is constant.Let the constant sum be $2a = 10 \ m$, so $a = 5 \ m$.The distance between the foci is $2ae = 8 \ m$, so $ae = 4 \ m$.Since $a = 5$, we have $5e = 4$, which gives $e = \frac{4}{5}$.Using the relation $b^2 = a^2(1 - e^2)$, we get $b^2 = 25(1 - (\frac{4}{5})^2) = 25(1 - \frac{16}{25}) = 25(\frac{9}{25}) = 9$.Thus, $b = 3 \ m$.The area of the ellipse is given by $\pi ab = \pi 	imes 5 	imes 3 = 15 \pi \ m^2$.

$A$ man running round a race-course notes that the sum of the distances of two flag-posts from him is always $10 \ m$ and the distance between the flag-posts is $8 \ m$. The area of the path he encloses in square metres is (in $\pi$)

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