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.
Best Answer
WLOG $P(t^2,t^3); Q(u^2,u^3), t\ne u$
$$\dfrac{dy}{dx}=\dfrac{3x^2}{2y}$$
So, the equation of the tangent at $P$ is $$\dfrac{y-t^3}{x-t^2}=\dfrac{3t^4}{2t^3}=\dfrac{3t}2$$
$$\iff x(3t)-2y-t^3=0$$
Now this passes through $Q$ $$\implies3tu^2-2u^3-t^3=0\iff\left(\dfrac tu\right)^3-3\left(\dfrac tu\right)+2=0$$
Clearly, one the roots is $-2$ What about the other roots?
Now $\dfrac{\tan\alpha}{\tan\beta}=\dfrac{3t}{2}\cdot\dfrac2{3u}=\dfrac tu$