9.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{2217be5e-8edd-413f-9c97-212e585ff58d-20_693_1038_267_450}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{figure}
A sketch of part of the curve \(C\) with equation
$$y = 20 - 4 x - \frac { 18 } { x } , \quad x > 0$$
is shown in Figure 3.
Point \(A\) lies on \(C\) and has \(x\) coordinate equal to 2
- Show that the equation of the normal to \(C\) at \(A\) is \(y = - 2 x + 7\).
The normal to \(C\) at \(A\) meets \(C\) again at the point \(B\), as shown in Figure 3 .
- Use algebra to find the coordinates of \(B\).
a) \(y = 20 - 4 ( 2 ) - \frac { 18 } { 2 } \quad \therefore \quad y = 3\)
$$\begin{aligned}
& y = 20 - 4 x - 18 x ^ { - 1 }
& \therefore \frac { d y } { d x } = - 4 + 18 x ^ { - 2 }
\end{aligned}$$ - \(x = 2 \quad \frac { d y } { d x } = - 4 + \frac { 18 } { 2 ^ { 2 } } = 1 / 2\)
$$m = - 2$$
\includegraphics[max width=\textwidth, alt={}]{2217be5e-8edd-413f-9c97-212e585ff58d-20_2258_50_313_1980}
9 continued
\includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-21_2253_51_315_35}
$$\begin{aligned}
\therefore \quad y & = - 2 x + c
3 & = - 2 ( 2 ) + c
c & = 7
y & = - 2 x + 7
\end{aligned}$$
b) \(20 - 4 x - \frac { 18 } { x } = - 2 x + 7\)
$$\begin{aligned}
\therefore & 13 - 2 x - \frac { 18 } { x } = 0
& 13 x - 2 x ^ { 2 } - 18 = 0
& 0 = 2 x ^ { 2 } - 13 x + 18
\therefore & x = \frac { - b \pm \sqrt { b ^ { 2 } - 4 a c } } { 2 a }
& a = 2 \quad b = - 13 \quad c = 18
& x = 2 \quad x = a / 2
\therefore & y = 3 \quad \therefore y = - 2
B = & ( a / 2 , - 2 )
\end{aligned}$$
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