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(A) Let $A$ and $B$ be the positions of the two flag posts and $P(x, y)$ be the position of the man. According to the problem,$PA + PB = 10$. We know

Vedclass · Accepted Answer

$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$

Vedclass · Answer

(A) Let $A$ and $B$ be the positions of the two flag posts and $P(x, y)$ be the position of the man. According to the problem,$PA + PB = 10$. We know that if a point moves in a plane such that the sum of its distances from two fixed points is constant,the path is an ellipse,and this constant value is equal to the length of the major axis $(2a)$. Therefore,the path is an ellipse with major axis $2a = 10 \, m$,so $a = 5$. The distance between the foci $(2c)$ is $8 \, m$,so $c = 4$. Using the relation $c^{2} = a^{2} - b^{2}$,we have $4^{2} = 5^{2} - b^{2}$. $16 = 25 - b^{2} \Rightarrow b^{2} = 9$. Thus,the equation of the ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,which is $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$.

$A$ man running on a racecourse notes that the sum of the distances from the two flag posts to him is always $10 \, m$ and the distance between the flag posts is $8 \, m$. Find the equation of the path traced by the man.

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