图 4 给出了曲线 $C$ 的草图，其参数方程为

(a) 用 $t$ 表示 $\frac{\mathrm{d}y}{\mathrm{d}x}$。

(b) 求当 $t=\dfrac{\pi}{3}$ 时曲线 $C$ 的切线方程。

答案写成 $y=mx+c$ 的形式，其中 $m,c$ 为常数。

(c) 证明曲线 $C$ 上所有点都满足方程

其中 $A,B$ 是待求常数。

分别对两个参数方程求导，再用参数导数相除求曲线斜率。

Differentiating with respect to $t$,

and

For a parametric curve,

Hence

也可以先展开正切和角公式，再用商法则求导。短记号 $N,D$ 可避免反复写长括号，最后恢复为 $t$ 的表达式。

The tangent addition formula gives

Put $u=\tan t$ and write

Then $y=N/D$, with

By the quotient rule,

Since $D+N=4$,

Also $\frac{\mathrm{d}u}{\mathrm{d}t}=\sec^2t$, so

With $\frac{\mathrm{d}x}{\mathrm{d}t}=\sec t\tan t$, this gives

where $D=1-\sqrt3\tan t$. Thus, entirely in terms of $t$,

This is equivalent to the expression in method 1.

代入给定参数求切点及两项参数导数，取它们的比得到斜率，最后写点斜式。

At $t=\frac{\pi}3$,

and

At this point, the two parameter derivatives are

and

Hence the tangent gradient is

Using $(2,-3)$,

先展开正切和角公式，再由 $x=\sec t$ 表示 $\tan t$。题设象限决定平方根取正；用短记号 $w$ 完成有理化。

Using the tangent addition formula,

Since $x=\sec t$,

The given range has $\tan t>0$, so

Put

Then

Rationalise by multiplying the numerator and denominator by $1+w$. The numerator is

which equals $3x^2+4w$ because $w^2=3x^2-3$. The denominator is

Thus

Comparing with the required form,

先用 $y$ 表示 $\tan t$，代入正割恒等式。取平方根前核对正负号，再解回 $y$ 并有理化。

By the tangent addition formula,

Rearranging,

so

Using $\sec^2t=1+\tan^2t$,

Hence

For the given parameter range,

Thus $y<-1$, so both $y-3$ and $y+1$ are negative. Their ratio is positive. Let $w=\sqrt{3x^2-3}$; then

Rearranging,

and therefore

Multiplying the numerator and denominator by $1+w$ gives

With $w^2=3x^2-3$, this becomes

Hence

也可以使用正弦、余弦。先展开和角公式，由 $x$ 表示正弦和余弦，再消去参数并有理化。

Write the tangent as sine divided by cosine. The compound-angle formulae give

and

Thus

Since $x=\sec t$,

Both sine and cosine are positive in the given range. Hence

Multiplying the numerator and denominator by $x$, and writing $w=\sqrt{3x^2-3}$, gives

Rationalising,

Substituting $w^2=3x^2-3$ gives

Therefore