8.

等轴双曲线 $H$ 的直角坐标方程为 $xy = 16$

点 $P\big(4t,\tfrac4t\big)$，$t \neq 0$，在 $H$ 上。

(a) 利用微积分证明 $H$ 在 $P$ 处的法线方程为

$H$ 上的点 $A$ 的参数为 $t = 2$

$H$ 在 $A$ 处的法线再次与 $H$ 相交于点 $B$。

(b) 求 $AB$ 的长度的精确值。

$H$ 在 $A$ 处的切线与 $y$ 轴相交于点 $C$。

(c) 求三角形 $ABC$ 的精确面积。

先把双曲线写成 $y=16x^{-1}$ 并求导。将点 $P$ 的横坐标 $x=4t$ 代入导数，求出切线斜率，再利用法线斜率是其负倒数写出点斜式方程。

Since

At $P$, where $x=4t$, the gradient of the tangent is

Therefore, the gradient of the normal is $t^2$. Using the point $P\left(4t,\dfrac{4}{t}\right)$, its equation is

Multiplying by $t$ and rearranging,

对 $xy=16$ 隐式求导，直接用点 $P$ 的两个坐标求切线斜率，再取负倒数写法线。

Differentiating $xy=16$ implicitly gives

At $P=(4t,4/t)$, where $t\ne0$,

Thus $m_N=t^2$. The normal through $P$ is

Multiplying by $t$,

分别对参数 $t$ 求导，再以纵坐标导数除以横坐标导数得到切线斜率。法线方程随后由点斜式得到。

For $x=4t$ and $y=4/t$,

Hence the tangent gradient is

Its negative reciprocal is $m_N=t^2$, so the normal at $P$ is

Since $t\ne0$, multiplying by $t$ and rearranging gives

先由 $t=2$ 求出 $A$，并将其代入 (a) 的法线方程。把法线与双曲线联立，其中一个交点已知是 $A$，另一个就是 $B$；得到两点坐标后使用距离公式。

For $t=2$,

From part (a), the normal at $A$ is

so

At an intersection with $H$,

and hence

The root $x=8$ corresponds to $A$, so at $B$,

Thus $B=\left(-\dfrac{1}{2},-32\right)$, and

由法线方程把 $x$ 写成 $y$ 的式子，再代入双曲线。解出两个纵坐标，排除已知交点 $A$ 后便得到 $B$。

At $t=2$, $A=(8,2)$ and the normal from part (a) is

Thus $x=(y+30)/4$. Substituting into $xy=16$,

The solution $y=2$ belongs to $A$. At $B$,

Therefore,

把一般参数点代入 $A$ 处的法线方程，得到关于参数的二次方程。排除 $t=2$，用另一个参数求出 $B$。

From part (a), the normal at $A=(8,2)$ is $y=4x-30$. A point $(4t,4/t)$ on this line satisfies

Multiplying by $t\ne0$ and dividing by $2$,

The parameter $t=2$ corresponds to $A$. Therefore, at $B$,

Hence

The distance is

先求切线在 $y$ 轴上的截距，从而得到 $C$。由于切线与法线在 $A$ 处垂直，三角形 $ABC$ 是以 $AB$、$AC$ 为直角边的直角三角形，可直接用两边长度求面积。

The gradient of the tangent at $A$ is $-\dfrac{1}{4}$, so its equation is

At the $y$-axis, $x=0$, giving $y=4$. Therefore,

Also,

Let $\mathcal A$ denote the area of triangle $ABC$. Since the tangent and normal at $A$ are perpendicular,

已知三个顶点坐标后，也可以直接使用坐标面积公式。这个方法不依赖切线与法线垂直的几何观察。

The tangent at $A$ is $y-2=-\frac14(x-8)$, so its $y$-intercept is $C=(0,4)$. Using

and denoting the area by $\mathcal A$, the coordinate area formula gives

法线与 $y$ 轴交于 $D$。由于 $A$、$B$ 在 $y$ 轴两侧，三角形 $ABC$ 可分成共用竖直底边 $CD$ 的两个三角形，两者的高分别为 $8$ 与 $1/2$。

From part (b),

The tangent at $A$ is

so $C=(0,4)$. The normal $AB$ has equation $y=4x-30$, and hence meets the $y$-axis at $D=(0,-30)$.

Thus $CD=34$. Since $A$ and $B$ are on opposite sides of the $y$-axis, the required area is the sum of the areas of triangles $ACD$ and $BCD$:

把三角形放进横坐标从 $-1/2$ 到 $8$、纵坐标从 $-32$ 到 $4$ 的矩形。用矩形面积减去三个角落的直角三角形面积。

From part (b), $A=(8,2)$ and $B=(-\frac12,-32)$. The tangent

gives $C=(0,4)$.

Introduce the rectangle vertices

The rectangle $BEFG$ has width $17/2$ and height $36$, so its area is $306$. The three corner triangles outside $ABC$ have areas

Here $[XYZ]$ denotes the area of triangle $XYZ$. Therefore,

也可以只补成梯形：下底长 $17/2$、上底长 $8$，高为 $36$。从梯形面积减去两侧的直角三角形面积即可。

From part (b), $A=(8,2)$ and $B=(-\frac12,-32)$. The tangent

gives $C=(0,4)$.

Let $E=(8,-32)$ and $F=(8,4)$. The parallel sides $BE$ and $CF$ of trapezium $BEFC$ have lengths $17/2$ and $8$, and the height is $36$. Thus its area is

Triangle $ABC$ is obtained by removing triangles $BEA$ and $CFA$. Hence