7. 抛物线 $C$ 的方程为 $y^2 = 4ax$，其中 $a$ 是正常数。

直线 $l$ 的方程为 $3x - 4y + 48 = 0$，是 $C$ 在点 $P$ 处的切线。

(a) 证明 $a = 9$

(b) 由此确定 $P$ 的坐标。

已知点 $S$ 是 $C$ 的焦点，直线 $l$ 与 $C$ 的准线相交于点 $A$，

(c) 求三角形 $PSA$ 的精确面积。

联立直线与抛物线，得到关于 $y$ 的二次方程。由于直线与抛物线相切，该方程有重根，因此判别式为零；再利用 $a>0$ 排除零解。

From the equation of the line,

Substituting this into $y^2=4ax$ gives

so

Since the line is tangent to the parabola, this quadratic has equal roots. Therefore,

Hence

Since $a>0$,

The same discriminant method can be applied after eliminating $y$. Substituting

into the parabola gives

A repeated root requires

Expanding and factorising gives

Again, $a>0$ gives $a=9$.

利用相切处斜率相等。对抛物线隐式求导，并与直线斜率 $\frac34$ 相等，先把切点坐标写成 $a$ 的表达式，再代回曲线和直线求出 $a$。

Differentiating $y^2=4ax$ implicitly gives

so

The line can be written as

so its gradient is $\frac34$. At the point of tangency,

which gives

Using $y^2=4ax$,

Substituting these coordinates into the equation of the line gives

Thus

and hence

使用抛物线的参数形式，分别对参数求导，得到切线斜率。将它与直线斜率相等，先求切点参数，再代入直线方程求 $a$。

Parameterise the parabola by

Then

At the point of tangency,

The tangent $l$ has gradient $\frac34$, so

and hence

Substituting the parametric coordinates into $l$ gives

Thus

Therefore,

so

承接 (a) 的联立方程，代入 $a=9$ 后求出重根 $y$，再代回直线方程求 $x$。

From part (a), the $y$-coordinate of the point of intersection satisfies

With $a=9$,

so

Therefore, $y=24$. Using $x=\frac{4y-48}{3}$,

Hence

Alternatively, the $x$-quadratic from part (a) becomes

that is,

Thus $x=16$, and $y=\frac34(16)+12=24$, giving the same point.

承接 (a) 的微分法，切点坐标已经用 $a$ 表示，因此只需代入 $a=9$。若使用参数微分法，也可直接代入已求出的切点参数。

From the differentiation method in part (a), the tangent point satisfies

Using $a=9$,

and

Hence

Equivalently, using $t=\frac43$ from the parametric method,

and

由 $y^2=4ax$ 的标准性质写出焦点和准线，再求出点 $A$。以 $S$ 为共同起点构造向量 $\overrightarrow{SP}$、$\overrightarrow{SA}$，用二维行列式的绝对值的一半求三角形面积。

Since $a=9$, the focus is

and the directrix is $x=-9$.

At $A$, $x=-9$. Substituting this into the equation of $l$ gives

so $y=\frac{21}{4}$. Hence

Using $P=(16,24)$,

and

The determinant formed by these two vectors is

Let $\mathcal A$ denote the area of $\triangle PSA$. Therefore,

The coordinate shoelace formula gives the same area. Taking the vertices in the order $S,A,P$, the signed double area is

Therefore,

在直线 $l$ 上取横坐标为 $9$ 的点 $Q$，则 $SQ$ 是竖直线段。线段 $SQ$ 把三角形 $PSA$ 分成两个三角形；它们具有共同的底 $SQ$，高分别是点 $P$、$A$ 到直线 $x=9$ 的水平距离。

As above,

and

Let $Q$ be the point on $l$ with $x=9$. Then

so

Thus $SQ=\frac{75}{4}$. The perpendicular distances from $P$ and $A$ to the line $x=9$ are respectively

and

Therefore,

Hence the exact area is

考虑切线与横轴之间、两条竖直线之间的梯形。它由所求三角形和底下两个三角形组成，因此用梯形面积减去这两个三角形面积。

As above,

and the directrix $x=-9$ meets $l$ at

Let $L=(-9,0)$ and $R=(16,0)$. The trapezium $LAPR$ has parallel sides $LA$ and $RP$, with

and perpendicular distance $LR=25$. Its area is

Since $LS=18$ and $SR=7$, the areas of the two lower triangles are

and

Hence the area of $\triangle PSA$ is

将图形放进边长为 $25$ 与 $24$ 的矩形中。扣除底下两个三角形，再扣除切线上方的三角形，便只剩所求三角形。

Use

and

Let

and $U=(-9,24)$. The rectangle $LRPU$ has area

The triangles below $AS$ and $SP$ have areas

and

For the triangle $UAP$ above the tangent,

and the perpendicular distance from $P$ to $UA$ is $25$. Its area is

Therefore, the required area is