<s>[INST] If the tangent at $P$ of the curve $y^2=x^3$ intersects the curve again at $Q$ If the tangent at $P$ of the curve $y^2=x^3$ intersects the curve again at $Q$ and the straight lines $OP,OQ$ make angles $\alpha,\beta$ with the $x$-axis where $O$ is the origin then $\frac{\tan\alpha}{\tan\beta}=$
$(A)-1$
$(B)-2$
$(C)2$
$(D)\sqrt2$

Let $P$ be $(x_1,y_1)$ and $Q$ be $(x_2,y_2)$.
Equation of tangent is $\frac{y-y_1}{x-x_1}=\frac{3x_1^2}{2y_1}$
It passes through $(x_2,y_2)$,so $\frac{y_2-y_1}{x_2-x_1}=\frac{3x_1^2}{2y_1}$
I need to find out $\frac{\tan\alpha}{\tan\beta}=\frac{\frac{y_1}{x_1}}{\frac{y_2}{x_2}}=\frac{y_1x_2}{x_1y_2}$
I am stuck here.
 [/INST] We have
\begin{align}
\frac{y_2-y_1}{x_2-x_1}&=\frac{3x_1^2}{2y_1}\\
2y_1y_2-2y_1^2&=3x_1^2x_2-3x_1^3\\
2y_1y_2-2y_1^2&=3x_1^2x_2-3y_1^2\\
2y_1y_2+y_1^2&=3x_1^2x_2\\
y_1^3(2y_2+y_1)^3&=27x_1^6x_2^3\\
y_1^3(2y_2+y_1)^3&=27y_1^4y_2^2\\
(2y_2+y_1)^3&=27y_1y_2^2\\
8y_2^3-15y_2^2y_1+6y_2y_1^2+y_1^3&=0\\
(y_2-y_1)^2(8y_2+y_1)&=0\\
8y_2&=-y_1\\
64y_2^2&=y_1^2\\
64x_2^3&=x_1^3\\
4x_2&=x_1
\end{align}
Therefore,
\begin{align}
\frac{\tan\alpha}{\tan\beta}&=\frac{y_1x_2}{x_1y_2}\\
&=\frac{-8y_2x_2}{4x_2y_2}\\
&=-2
\end{align}
This method works as when we solve the equation of the tangent with the curve, the $y$-coordinate (or $x$-coordinate if we rewrite it as an equation in $x$) of $P$ is a double root of the cubic equation. It is easy to find the third root, which gives $Q$.
</s>