Question

If the tangent at $(x_1,y_1)$ to the curve $x^3+y^3=a^3$ meets the curve again at $(x_2,y_2)$ then

• A. $\frac{x_2}{x_1}+\frac{y_2}{y_1}=-1$
• B. $\frac{x_2}{y_1}+\frac{x_1}{y_2}=-1$
• C. $\frac{x_1}{x_2}+\frac{y_1}{y_2}=-1$
• D. $\frac{x_2}{x_1}+\frac{y_2}{y_1}=1$

Answer 1

The correct option is A $\frac{x_2}{x_1}+\frac{y_2}{y_1}=-1$

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

The derivative is

$\frac{dy}{dx}=-\frac{x^2}{y^2}.....\left(1\right)$

Therefore, slope of tangent at $(x_1,y_1)$ is

$-\frac{x_1^2}{y_1^2}.........\left(2\right)$

The tangent passes through $(x_2,y_2)$, therefore the slope of tangent is also given by

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

Comparing the two slope equations we get

$\frac{y_2-y_1}{x_2-x_1}=-\frac{x_1^2}{y_1^2}$

$\frac{y_2^3-y_1^3}{x_2^3-x_1^3}\times \frac{x_1^2+x_1{x}_2+x_2^2}{y_1^2+y_1{y}_2+y_2^2}=-\frac{x_1^2}{y_1^2}$

$-\frac{x_1^2+x_1{x}_2+x_2^2}{y_1^2+y_1{y}_2+y_2^2}=-\frac{x_1^2}{y_1^2}$

$x_1^2{y}_1^2+x_1{x}_2{y}_1^2+x_2^2{y}_1^2=x_1^2{y}_1^2+x_1^2{y}_1{y}_2+x_1^2{y}_2^2$

$x_1{x}_2{y}_1^2+x_2^2{y}_1^2=x_1^2{y}_1{y}_2+x_1^2{y}_2^2$

$x_1^2{y}_2^2-x_2^2{y}_1^2=x_1{x}_2{y}_1^2-x_1^2{y}_1{y}_2$

$(x_1{y}_2-x_2{y}_1)(x_1{y}_2+x_2{y}_1)=x_1{y}_1(x_2{y}_1-x_1{y}_2)$

$x_1{y}_2+x_2{y}_1=-x_1{y}_1$

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