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{d y}{d x}\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 $(\alpha, \beta) \&\left(\alpha_1, \beta_1\right)$
$\Rightarrow-\frac{\alpha^2}{\beta^2}=\frac{\beta_1-\beta}{\alpha_1-\alpha} \& \alpha^3+\beta^3=\alpha_1^3+\beta_1^3$
$\Rightarrow \beta_1^3-\beta^3=\alpha^3-\alpha_1^3$
$\Rightarrow\left(\beta_1-\beta\right)\left(\beta^2+\beta \beta_1+\beta_1^2\right)=-\left(\alpha_1-\alpha\right)\left(\alpha^2+\alpha \alpha_1+\alpha_1^2\right)$
$\therefore+\frac{\alpha^2}{\beta^2}=+\frac{\alpha^2+\alpha \alpha_1+\alpha_1^2}{\beta^2+\beta \beta_1+\beta_1^2}$
$\Rightarrow \frac{\alpha_1}{\alpha}+\frac{\beta_1}{\beta}=-1$