2016.3.2 Question 2

1. For $y^2=4\mathit{ax}$, we have $x=\frac{y^2}{4a}$, and therefore

   $$\frac{dx}{dy}=\frac{2y}{4a}.$$

   Therefore, the normal through $Q$, $l_Q$ satisfies that

   $$l_Q:x-a{q}^2=-\frac{4a}{2\cdot 2\mathit{aq}}\cdot \left(y-2\mathit{aq}\right),$$

   i.e.

   $$l_Q:q(x-a{q}^2)=-\left(y-2\mathit{aq}\right).$$

   Since $P\in l_Q$, we must have

   $$\begin{array}{llll}\hfill q(a{p}^2-a{q}^2)& =-\left(2\mathit{ap}-2\mathit{aq}\right)& \hfill & \\ \hfill \mathit{aq}(p+q)(p-q)& =-2a(p-q)& \hfill & \\ \hfill \mathit{pq}+q^2& =-2& \hfill & \\ \hfill q^2+\mathit{pq}+2& =0& \hfill & \end{array}$$

   as desired.

2. We also have

   $$r^2+\mathit{pr}+2=0.$$

   Since $q\ne r$, $q,r$ are the solutions to the equation

   $$x^2+\mathit{px}+2=0,$$

   and therefore $q+r=-p,\mathit{qr}=2$.

   Note that the equation for $\mathit{QR}$ satisfies that

   $$m_{\mathit{QR}}=\frac{2\mathit{ar}-2\mathit{aq}}{a{r}^2-a{q}^2}=\frac{2}{r+q}.$$

   Therefore, $l_{\mathit{QR}}$ satisfies that

   $$\begin{array}{llll}\hfill l_{\mathit{QR}}:y-2\mathit{aq}& =\frac{2}{r+q}(x-a{q}^2)& \hfill & \\ \hfill y& =\frac{2}{r+q}\left(x-a{q}^2+\frac{r+q}{2}\cdot 2\mathit{aq}\right)& \hfill & \\ \hfill y& =\frac{2}{r+q}\left(x-a{q}^2+a{q}^2+\mathit{aqr}\right)& \hfill & \\ \hfill y& =\frac{2}{r+q}\left(x+\mathit{aqr}\right)& \hfill & \\ \hfill y& =-\frac{2}{p}(x+2a).& \hfill & \end{array}$$

   This passes through a fixed point $(-2a,0)$.

3. $\mathit{OP}$ has equation $y=\frac{2\mathit{ap}}{a{p}^2}x$, which is $y=\frac{2x}{p}$. Therefore, since $T=\mathit{OP}\cap \mathit{QR}$, $x_T$ must satisfy that

   $$\begin{array}{llll}\hfill -\frac{2}{p}(x+2a)& =\frac{2x}{p},& \hfill & \\ \hfill -(x+2a)& =x& \hfill & \\ \hfill x& =-a.& \hfill & \end{array}$$

   Therefore, $y_T=-\frac{2a}{p}$, $T\left(-a,-\frac{2a}{p}\right)$ lies on the line $x=-a$ which is independent of $p$.

   The distance from the $x$-axis to $T$ is $\left|\frac{2a}{p}\right|=\frac{2a}{\left|p\right|}$.

   Notice that since $\mathit{qr}=2$, $q$ and $r$ must take the same parity, and therefore $\left|p\right|=\left|q\right|+\left|r\right|$. By the AM-GM inequality, we have

   $$\left|q\right|+\left|r\right|\ge 2\sqrt{\left|q\right|\cdot \left|r\right|}=2\sqrt{2},$$

   with the equal sign holding if and only if $\left|q\right|=\left|r\right|$, $q=r$, which is impossible.

   Therefore, $\left|p\right|>2\sqrt{2}$ and therefore $\frac{2a}{\left|p\right|}<\sqrt{2}$ as desired.