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 :

• A. $\left(\frac{x_1}{x_2}\right)+\left(\frac{y_1}{y_2}\right)=0$
• B. $\left(\frac{x_2}{x_1}\right)+\left(\frac{y_2}{y_1}\right)=0$
• C. $\frac{x_1}{x_2}+\frac{y_1}{y_2}+1=0$
• D. $\frac{x_2}{x_1}+\frac{y_2}{y_1}+1=0$

Answer 1

The correct option is D

$\frac{x_2}{x_1}+\frac{y_2}{y_1}+1=0$

Find the relation

Given, $x^3+y^3=a^3.$

Points $(x_1,y_1)$ and $(x_2,y_2)$ lies on the curve

${x_1}^3+{y_1}^3=a^3......\left(1\right)$

${x_2}^3+{y_2}^3=a^3.....\left(2\right)$

From $\left(2\right)-\left(1\right)$

${x_2}^3-{x_1}^3=-({y_2}^3-{y_1}^3)......\left(3\right)$

The slope of the tangent $=\frac{dy}{dx}=-\left(\frac{x^2}{y^2}\right)$

The slope at $(x_1,y_1)$ $=-\left(\frac{{x_1}^2}{{y_1}^2}\right)$

The equation of the tangent:

$\left(y-y_1\right)=-\left(\frac{{x_1}^2}{{y_1}^2}\right)(x-x_1)$

The equation of the tangent at the point $(x_2,y_2)$ is:

$$\begin{gathered} =\left(y_2-y_1\right)=-\left(\frac{{x_1}^2}{{y_1}^2}\right)(x_2-x_1) \\ \Rightarrow -{y_1}^2\left(y_2-y_1\right)={x_1}^2(x_2-x_1) \end{gathered}$$

From $\left(3\right)$

$\Rightarrow \frac{x_2}{x_1}+\frac{y_2}{y_1}+1=0$

Hence, the correct option is $D$.