Question

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 :
$(1)\left( {\dfrac{{{x_1}}}{{{x_2}}}} \right) + \left( {\dfrac{{{y_1}}}{{{y_2}}}} \right) = 0$
$(2)\left( {\dfrac{{{x_2}}}{{{x_1}}}} \right) + \left( {\dfrac{{{y_2}}}{{{y_1}}}} \right) = 0$
$(3)\left( {\dfrac{{{x_1}}}{{{x_2}}}} \right) + \left( {\dfrac{{{y_1}}}{{{y_2}}}} \right) + 1 = 0$
$(4)\left( {\dfrac{{{x_2}}}{{{x_1}}}} \right) + \left( {\dfrac{{{y_2}}}{{{y_1}}}} \right) + 1 = 0$

Answer 1

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.