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.

Simplify the expression a bit.
Notice that tangent of an angle is just the slope it makes.
So, we need to find the ratio of the slopes.
Let $P$ be $(1,1)$.
We have the derivative of $y=\displaystyle \sqrt{x^3} = \frac32\sqrt{x}$ .
Therefore, the tangent line is $(y-1)=\frac{3}{2}(x-1)$.
Because $y^2=x^3$ is concave upward for $y>0$, we know that the line will intersect the function again at a Q such that $y<0$.
Solving for Q, we have $\displaystyle -x^{\frac{3}{2}}-1=\frac{3}{2}x-\frac{3}{2}$, or $x^{\frac32}+\frac{3}{2}x-\frac{1}{2}=0$.
We simplify as $\frac{1}{2}(\sqrt{x}+1)^2(2\sqrt{x}-1)=0$.
Because we have $\sqrt{x} \neq 1$, we must have $2\sqrt{x}=1$,and therefore $x=\frac{1}{4}$, and $Q=\displaystyle \left(\frac14, -\frac18\right)$.
We have the slope of $OP$ is $1$ and the slope of $OQ$ is $\displaystyle -\frac12$.
Therefore, finally, $\displaystyle \frac{\tan \alpha}{\tan \beta}=-2$, or (B)
