5. 等轴双曲线 $H$ 的方程为 $xy = 64$

点 $P\left(8p, \dfrac{8}{p}\right)$，其中 $p \neq 0$，在 $H$ 上。

(a) 利用微积分证明 $H$ 在 $P$ 处的法线方程为

$H$ 在 $P$ 处的法线再次与 $H$ 相交于点 $Q$。

(b) 用 $p$ 表示 $Q$ 的坐标，答案化为最简形式。

把双曲线写成 $y=64x^{-1}$ 后求导，得到切线斜率。代入点 $P$ 的横坐标，再利用法线斜率是切线斜率的负倒数，最后用点斜式推导题目指定的方程。

From $xy=64$,

Differentiating with respect to $x$ gives

At $P\left(8p,\frac{8}{p}\right)$, the gradient of the tangent is

Therefore, the gradient of the normal is $p^2$. Its equation is

Multiplying by $p$ and rearranging,

Hence the normal has equation

直接对原方程隐式求导。用乘积法则求出切线斜率，并代入点的两个坐标；再取负倒数得到法线斜率，用点斜式整理成指定形式。

Differentiating $xy=64$ implicitly gives

so

At $P$, where $p\ne0$,

The normal gradient is therefore

Its equation is

Multiplying by $p$ and rearranging,

Thus the normal has equation

将曲线写成参数形式，分别对参数求导，再用两导数相除得到切线斜率。点所对应的参数为 $p$，代入后求法线方程。

Parameterise the hyperbola by

where $t\ne0$. Then

and

Hence

At $P$, $t=p$, so the tangent gradient is $-p^{-2}$ and the normal gradient is $p^2$.

Therefore,

Multiplying by $p$ gives

and rearranging gives

联立法线与双曲线，用 $y=\frac{64}{x}$ 消去 $y$，得到关于 $x$ 的二次方程。已知其中一个根对应点 $P$，因式分解后取另一个根，再由 $xy=64$ 求出 $Q$ 的纵坐标。

Since $Q$ lies on $H$,

Substituting this into the equation of the normal gives

Multiplying by $x$ and rearranging,

This factorises as

The root $x=8p$ corresponds to $P$, so at the second point of intersection,

Hence

Therefore,

把双曲线上任意点参数化为 $\left(8q,\frac{8}{q}\right)$。将它代入法线方程，得到关于 $q$ 的二次方程；其中 $q=p$ 对应已知点 $P$，另一个根就给出点 $Q$。

Write a point on $H$ as

Substituting this into the normal equation gives

Multiplying by $q$ and rearranging,

Equivalently,

The solution $q=p$ corresponds to $P$. For the second point of intersection,

Therefore,

沿用关于横坐标的二次方程，但用根与系数关系代替因式分解。已知一根是点 $P$ 的横坐标，用两根之和减去它，便得到点 $Q$ 的横坐标。

As in the first method, the intersection coordinates satisfy

One root is $8p$, the $x$-coordinate of $P$. Let the other root be $x_Q$. Since $p\ne0$, the sum of the roots is

Thus

Using $xy=64$,

Therefore,

改为消去横坐标，得到关于纵坐标的二次方程。其中一个根对应已知点 $P$，因式分解后取另一个根，再由双曲线方程求横坐标。

Since $xy=64$, write

Substituting into the normal equation gives

Multiplying by $y$ and rearranging,

Factorising,

The root $y=\frac8p$ corresponds to $P$. Therefore, at $Q$,

and hence

Thus