# 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.

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 $Q$.

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 $\dfrac{\tan\alpha}{\tan\beta}=\dfrac{3t}{2}\cdot\dfrac2{3u}=\dfrac tu$

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)

• Curious as to why this was down voted. – Saketh Malyala Jun 18 '17 at 6:13