题目

Problem

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

The curve $C$ has equation

\begin{align*} y=\frac23x^3-25x-\frac{56}{x}+\frac{194}{3}, \qquad x>0. \end{align*}

The point $P$ , which lies on $C$ , has coordinates $(2,-8)$ .

(a) Show that an equation of the tangent to $C$ at $P$ is

\begin{align*} y=-3x-2. \end{align*}

(5)

The point $Q$ also lies on $C$ .

Given that the tangent to $C$ at $Q$ is parallel to the tangent to $C$ at $P$ ,

(b) find, using algebra and showing your working, the exact $x$ coordinate of $Q$ .

(5)

解答

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先求导，再把 $x=2$ 代入求切线斜率。切线经过 $P(2,-8)$ ，用点斜式即可推出目标直线。

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Differentiate:

\begin{align*} \frac{dy}{dx} =&\,2x^2-25+\frac{56}{x^2}. \end{align*}

At $P$ , $x=2$ , so

\begin{align*} \left.\frac{dy}{dx}\right|_{x=2} =&\,2(2)^2-25+\frac{56}{2^2}\\ =&\,8-25+14\\ =&\,-3. \end{align*}

The tangent at $P(2,-8)$ has gradient $-3$ , so

\begin{align*} y-(-8)=&\,-3(x-2)\\ y+8=&\,-3x+6\\ y=&\,-3x-2. \end{align*}

This is the required result.

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平行切线说明斜率相同，所以令导数等于 $-3$ 。整理时会得到关于 $x^2$ 的二次方程；一个解对应原来的点 $P$ ，另一个正解就是点 $Q$ 的 $x$ 坐标。

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For a tangent parallel to the tangent at $P$ ,

\begin{align*} \frac{dy}{dx}=-3. \end{align*}

\begin{align*} 2x^2-25+\frac{56}{x^2}=&\,-3\\ 2x^2-22+\frac{56}{x^2}=&\,0. \end{align*}

Multiply by $x^2$ :

\begin{align*} 2x^4-22x^2+56=&\,0\\ x^4-11x^2+28=&\,0. \end{align*}

Let $u=x^2$ . Then

\begin{align*} u^2-11u+28=&\,0\\ (u-4)(u-7)=&\,0. \end{align*}

Therefore

\begin{align*} x^2=4 \quad\text{or}\quad x^2=7. \end{align*}

Since $x>0$ ,

\begin{align*} x=2 \quad\text{or}\quad x=\sqrt7. \end{align*}

The value $x=2$ gives the point $P$ , so the exact $x$ coordinate of $Q$ is

\begin{align*} \sqrt7. \end{align*}