Two men are walking on a path $ {x^3} + {y^3} = {a^3} $ . When the first man arrives at a point $ ({x_1},{y_1}) $ , he finds the second man in the direction of his own instantaneous motion. If the coordinates of the second man are $ ({x_2},{y_2}) $ then :
 $ (1)\left( {\dfrac{{{x_1}}}{{{x_2}}}} \right) + \left( {\dfrac{{{y_1}}}{{{y_2}}}} \right) = 0 $
 $ (2)\left( {\dfrac{{{x_2}}}{{{x_1}}}} \right) + \left( {\dfrac{{{y_2}}}{{{y_1}}}} \right) = 0 $
 $ (3)\left( {\dfrac{{{x_1}}}{{{x_2}}}} \right) + \left( {\dfrac{{{y_1}}}{{{y_2}}}} \right) + 1 = 0 $
 $ (4)\left( {\dfrac{{{x_2}}}{{{x_1}}}} \right) + \left( {\dfrac{{{y_2}}}{{{y_1}}}} \right) + 1 = 0 $

Hint: This question is based on the concept that if a point lies on a curve, then the coordinates of the point will satisfy the equation of the curve. The two important identities which will be used to solve this question are:
 $ ({a^3} - {b^3}) = (a - b)({a^2} + {b^2} + ab) $
 $ {a^2} - {b^2} = (a + b)(a - b) $ .

Complete answer:
Since both the point $ ({x_1},{y_1}) $ and $ ({x_2},{y_2}) $ lie on the same curve $ {x^3} + {y^3} = {a^3} $ , so,
 $ x_1^3 + y_1^3 = {a^3}........(1) $
Similarly,
 $ x_2^3 + y_2^3 = {a^3}........(2) $
On subtracting equation (1) from equation (2),
 $ (x_2^3 - x_1^3) + (y_2^3 - y_1^3) = 0 $
 $ (x_2^3 - x_1^3) = - (y_2^3 - y_1^3).......(3) $
We know that the equation of the path is $ {x^3} + {y^3} = {a^3} $
On differentiating this equation,
 $ 3{x^2} + 3{y^2}\left( {\dfrac{{dy}}{{dx}}} \right) = 0 $
 $ 3{y^2}\left( {\dfrac{{dy}}{{dx}}} \right) = - 3{x^2} $
On cancelling $ 3 $ on both the sides,
 $ {y^2}\left( {\dfrac{{dy}}{{dx}}} \right) = - {x^2} $
 $ \left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{ - {x^2}}}{{{y^2}}} $
Hence, we can say that the slope of the tangent at $ ({x_1},{y_1}) $ is $ \left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{{ - {x^2}}}{{{y^2}}} $
The equation of the tangent at $ ({x_1},{y_1}) $ is $ y - {y_1} = - \left( {\dfrac{{x_1^2}}{{y_1^2}}} \right)(x - {x_1}) $
Now this passes through $ ({x_2},{y_2}) $
 $ {y_2} - {y_1} = - \left( {\dfrac{{x_1^2}}{{y_1^2}}} \right)({x_2} - {x_1}) $
On cross multiplying, we get,
 $ \left( {\dfrac{{{x_2}}}{{{x_1}}} + \dfrac{{{y_2}}}{{{y_1}}} + 1} \right) = 0 $
On dividing equation (3) and (4), we get,
 $ \dfrac{{(x_2^3 - x_1^3)}}{{x_1^2({x_2} - {x_1})}} = \dfrac{{ - (y_2^3 - y_1^3)}}{{ - y_1^2({y_2} - {y_1})}} $
On putting the identity $ ({a^3} - {b^3}) = (a - b)({a^2} + {b^2} + ab) $ ,
 $ \dfrac{{({x_2} - {x_1})(x_2^2 + x_1^2 + {x_1}{x_2})}}{{x_1^2({x_2} - {x_1})}} = \dfrac{{ - ({y_2} - {y_1})(y_2^2 + y_1^2 + {y_1}{y_2})}}{{ - y_1^2({y_2} - {y_1})}} $
 $ \dfrac{{(x_2^2 + x_1^2 + {x_1}{x_2})}}{{x_1^2}} = \dfrac{{(y_2^2 + y_1^2 + {y_1}{y_2})}}{{y_1^2}} $
On further simplifying, we get,
 $ {\left( {\dfrac{{{x_2}}}{{{x_1}}}} \right)^2} + 1 + \left( {\dfrac{{{x_2}}}{{{x_1}}}} \right) = {\left( {\dfrac{{{y_2}}}{{{y_1}}}} \right)^2} + 1 + \left( {\dfrac{{{y_2}}}{{{y_1}}}} \right) $
 $ {\left( {\dfrac{{{x_2}}}{{{x_1}}}} \right)^2} + \left( {\dfrac{{{x_2}}}{{{x_1}}}} \right) = {\left( {\dfrac{{{y_2}}}{{{y_1}}}} \right)^2} + \left( {\dfrac{{{y_2}}}{{{y_1}}}} \right) $
On taking all the terms on the same side,
 $ {\left( {\dfrac{{{x_2}}}{{{x_1}}}} \right)^2} - {\left( {\dfrac{{{y_2}}}{{{y_1}}}} \right)^2} + \left( {\dfrac{{{x_2}}}{{{x_1}}}} \right) - \left( {\dfrac{{{y_2}}}{{{y_1}}}} \right) = 0 $
On applying the identity $ {a^2} - {b^2} = (a + b)(a - b) $ on $ {\left( {\dfrac{{{x_2}}}{{{x_1}}}} \right)^2} - {\left( {\dfrac{{{y_2}}}{{{y_1}}}} \right)^2} $ , we get,
 $ \left( {\dfrac{{{x_2}}}{{{x_1}}} + \dfrac{{{y_2}}}{{{y_1}}}} \right)\left( {\dfrac{{{x_2}}}{{{x_1}}} - \dfrac{{{y_2}}}{{{y_1}}}} \right) + \left( {\dfrac{{{x_2}}}{{{x_1}}} - \dfrac{{{y_2}}}{{{y_1}}}} \right) = 0 $
On taking $ \left( {\dfrac{{{x_2}}}{{{x_1}}} - \dfrac{{{y_2}}}{{{y_1}}}} \right) $ as common,
 $ \left( {\dfrac{{{x_2}}}{{{x_1}}} - \dfrac{{{y_2}}}{{{y_1}}}} \right)\left( {\dfrac{{{x_2}}}{{{x_1}}} + \dfrac{{{y_2}}}{{{y_1}}} + 1} \right) = 0 $
So, we can say that \[\left( {\dfrac{{{x_2}}}{{{x_1}}}} \right) + \left( {\dfrac{{{y_2}}}{{{y_1}}}} \right) + 1 = 0\]
So, the correct answer is $ (4)\left( {\dfrac{{{x_2}}}{{{x_1}}}} \right) + \left( {\dfrac{{{y_2}}}{{{y_1}}}} \right) + 1 = 0 $

Note:
The coordinates of a point are a pair of numbers that define its exact location on a two-dimensional plane. The coordinate plane has two axes at right angles to each other, called the $ x $ and $ y $ axis. The coordinates of a given point represent how far along each axis the point is located.