2020-M2-07

Since $\Gamma$ passes through the point $(1,2)$, we have

$\begin{array}{rcl}
2 & = & -(1)^2+8(1)+C \\
C & = & -5
\end{array}$

Therefore, the equation of $\Gamma$ is $y=-x^2+8x-5$.

1. Let $(h,k)$ be the coordinates of $P$. Hence by the result of (a), we have $k=-h^2+8h-5 \ \ldots \unicode{x2460}$.

   Since $L$ passes through $(5,14)$, by the slope formula, we have

   $\begin{array}{rcl}
   m_L & = & \dfrac{k-14}{h-5}
   \end{array}$

   Since the slope of the tangent of $\Gamma$ at any point is $f'(x)=-2x+8$, at $P(h,k)$, we have

   $\begin{array}{rcl}
   f'(h) & = & -2h+8
   \end{array}$

   Hence, we have

   $\begin{array}{rcl}
   -2h+8 & = & \dfrac{k-14}{h-5} \\
   -2h^2+18h-40 & = & k-14\\
   k & = & -2h^2+18h-26 \ \ldots \unicode{x2461}
   \end{array}$

   Sub. $\unicode{x2460}$ into $\unicode{x2461}$, we have

   $\begin{array}{rcl}
   -h^2+8h-5 & = & -2h^2+18h-26 \\
   h^2-10h+21 & = & 0 \\
   (h-3)(h-7) & = & 0
   \end{array}$

   $\therefore h=3$ or $h=7$.

   Note that the slope of $L$ is negative.

   Sub. $h=3$ into $f'(x)=-2x+8$, we have

   $\begin{array}{rcl}
   f'(3) & = & -2(3)+8 \\
   f'(3) & = & 2 \\
   f'(3) & > & 0
   \end{array}$

   Sub. $h=7$ into $f'(x)=-2x+8$, we have

   $\begin{array}{rcl}
   f'(7) & = & -2(7)+8 \\
   f'(7) & = & -6 \\
   f'(7) & < & 0 \end{array}$

   Therefore, the $x$-coordinate of $P$ is $7$.

   Sub. $h=7$ into $\unicode{x2460}$, we have

   $\begin{array}{rcl}
   k & = & =-(7)^2+8(7)-5 \\
   k & = & 2
   \end{array}$

   Hence, the coordinates of $P$ are $(7,2)$.

2. This is out of syllabus after 2022.

   The slope of the normal to $\Gamma$ at $P$

   $\begin{array}{cl}
   = & -1 \div m_L \\
   = & -1 \div \dfrac{2-14}{7-5} \\
   = & -1 \div (-6) \\
   = & \dfrac{1}{6}
   \end{array}$

   Therefore, the equation of the normal to $\Gamma$ at $P$ is

   $\begin{array}{rcll}
   \dfrac{y-2}{x-7} & = & \dfrac{1}{6} \\
   6y-12 & = & x-7 \\
   x-6y+5 & = & 0
   \end{array}$