Let A and B be the positions of the two flag posts and P(x, y) be the position of the man.

So, PA + PB = 10.

We know that if a point moves in plane in such a way that the sum of its distance from two fixed point is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Then, the path described by the man is an ellipse where the length of the major axis is 10m, while points A and B are the foci.

Now let us take the origin of the coordinate plane as the centre of the ellipse, and taking the major axis along the x- axis,

The equation of the ellipse is in the form of \( \dfrac{ x^2}{a^2} + \dfrac{y^2}{b^2} = 1\), where ‘a’ is the semi-major axis.

So, 2a = 10

a = \( \dfrac{10}{2}\)

= 5

Distance between the foci, 2c = 8

c = \( \dfrac{8}{2}\)

= 4

By using the relation, \( c=\sqrt{(a^2 – b^2)}\), we get,

4 = \( \sqrt{(25 – b^2)}\)

16 = 25 – b^{2}

b^{2} = 25 -1

= 9

b = 3

Hence, equation of the path traced by the man is \(\dfrac{ x^2}{25} + \dfrac{y^2}{9} = 1\)

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