Question

If tangent at $(\alpha ,\beta )$ to the curve $x^3+y^3=c^3$ meets curve again in $\left({\alpha }_1,{\beta }_1\right)$ then $\frac{{\alpha }_1}{\alpha }+\frac{{\beta }_1}{\beta }=$

• A. $-1$
• B. 1
• C. 0
• D. 2

Answer 1

Answer: (A)

$\because x^3+y^3=c^3$
$\because \left(\frac{dy}{dx}\right)$ at $(\alpha ,\beta )=\frac{-{\alpha }^2}{{\beta }^2}\&{\alpha }^3+{\beta }^3=c^3,{\alpha }_1{}^3+{\beta }_1{}^3=c^3$
slope of tangent $=$ slope of line joining $(\alpha ,\beta )\&\left({\alpha }_1,{\beta }_1\right)$
$\Rightarrow -\frac{{\alpha }^2}{{\beta }^2}=\frac{{\beta }_1-\beta }{{\alpha }_1-\alpha }\&{\alpha }^3+{\beta }^3={\alpha }_1^3+{\beta }_1^3$
$\Rightarrow {\beta }_1^3-{\beta }^3={\alpha }^3-{\alpha }_1^3$
$\Rightarrow \left({\beta }_1-\beta \right)\left({\beta }^2+\beta {\beta }_1+{\beta }_1^2\right)=-\left({\alpha }_1-\alpha \right)\left({\alpha }^2+\alpha {\alpha }_1+{\alpha }_1^2\right)$
$\therefore +\frac{{\alpha }^2}{{\beta }^2}=+\frac{{\alpha }^2+\alpha {\alpha }_1+{\alpha }_1^2}{{\beta }^2+\beta {\beta }_1+{\beta }_1^2}$
$\Rightarrow \frac{{\alpha }_1}{\alpha }+\frac{{\beta }_1}{\beta }=-1$