题目

Problem

The parabola $C$ has equation $y^2 = \dfrac{1}{2}x$

The point $P\left(\dfrac{t^2}{8}, \dfrac{t}{4}\right)$ , where $t \neq 0$ , lies on $C$ .

(a) Use calculus to show that the tangent to $C$ at $P$ has equation

8yt - 8x = t^2

(3)

Given that

the tangent to $C$ at $P$ meets the $y$ -axis at the point $A$
the line $l_1$ is the perpendicular bisector of the line segment $OA$ where $O$ is the origin
the line $l_2$ is the perpendicular bisector of the line segment $AP$
$l_1$ and $l_2$ intersect at the point $Q$

(b) show that, as $t$ varies, the coordinates of $Q$ satisfy the equation

y^2 = \alpha x + \beta

where $\alpha$ and $\beta$ are constants to be determined.

(5)

题目中文翻译

抛物线 $C$ 的方程为 $y^2 = \dfrac{1}{2}x$ 。点 $P\left(\dfrac{t^2}{8}, \dfrac{t}{4}\right)$ （其中 $t \neq 0$ ）位于 $C$ 上。

(a) 利用微积分学证明，抛物线 $C$ 在点 $P$ 处的切线方程为

8yt - 8x = t^2

(3)

已知：

在点 $P$ 处的切线与 $y$ 轴相交于点 $A$
直线 $l_1$ 是线段 $OA$ 的垂直平分线，其中 $O$ 为原点
直线 $l_2$ 是线段 $AP$ 的垂直平分线
$l_1$ 与 $l_2$ 相交于点 $Q$

(b) 证明：当 $t$ 变化时，点 $Q$ 的坐标满足方程

y^2 = \alpha x + \beta

其中 $\alpha$ 和 $\beta$ 为待求的常数。

(5)

解答

(a)

解法一

思路

展开

对抛物线方程作隐式求导，得到切线斜率；再代入点 $P$ 的纵坐标，把斜率化为 $\frac1t$ 。最后用点斜式并整理到题目指定的形式。

答题过程

展开

Differentiating

y^2=\frac12x

implicitly with respect to $x$ gives

2y\frac{\mathrm{d}y}{\mathrm{d}x}=\frac12.

Hence

\frac{\mathrm{d}y}{\mathrm{d}x}=\frac1{4y}.

At $P\big(\frac{t^2}{8},\frac t4\big)$ ,

\frac{\mathrm{d}y}{\mathrm{d}x} =\frac1{4(t/4)} =\frac1t.

Therefore, the tangent at $P$ is

\begin{align*} y-\frac t4 =&\,\frac1t\big(x-\frac{t^2}{8}\big)\\ 8ty-2t^2 =&\,8x-t^2\\ 8ty-8x =&\,t^2, \end{align*}

as required.

(b)

解法一

思路

展开

先由切线的 $y$ 轴截距求出 $A$ 。由于 $OA$ 是竖直线段，其垂直平分线 $l_1$ 是水平线，可立即得到 $Q$ 的纵坐标。再求 $AP$ 的中点和斜率，写出其垂直平分线 $l_2$ ；联立 $l_1,l_2$ 得到 $Q$ 的参数坐标，最后用 $t=16y$ 消去参数。

答题过程

展开

At the $y$ -axis, $x=0$ . From the tangent equation,

8ty=t^2.

Since $t\ne0$ ,

A=\big(0,\frac t8\big).

The midpoint of $OA$ is $\big(0,\frac t{16}\big)$ . Since $OA$ is vertical, its perpendicular bisector $l_1$ is

y=\frac t{16}.

Thus, for the point $Q$ ,

y_Q=\frac t{16}.

The coordinates of the midpoint of $AP$ are

\begin{align*} x_M =&\,\frac{0+t^2/8}{2}\\ =&\,\frac{t^2}{16},\\ y_M =&\,\frac{t/8+t/4}{2}\\ =&\,\frac{3t}{16}. \end{align*}

Thus

M=\big(\tfrac{t^2}{16},\tfrac{3t}{16}\big).

Also, the gradient of $AP$ is

\frac{t/4-t/8}{t^2/8-0}=\frac1t.

Therefore, the gradient of its perpendicular bisector $l_2$ is $-t$ , so

y-\frac{3t}{16} =-t\big(x-\frac{t^2}{16}\big).

At $Q$ , substitute $y=\frac t{16}$ . Since $t\ne0$ , we may divide by $t$ in the following working:

\begin{align*} \frac t{16}-\frac{3t}{16} =&\,-t\big(x-\frac{t^2}{16}\big)\\ -\frac{2t}{16} =&\,-t\big(x-\frac{t^2}{16}\big)\\ -\frac18 =&\,-x+\frac{t^2}{16}\\ x =&\,\frac{t^2+2}{16}. \end{align*}

Since $y=\frac t{16}$ , we have $t=16y$ . Hence

\begin{align*} x =&\,\frac{(16y)^2+2}{16}\\ =&\,16y^2+\frac18. \end{align*}

Therefore,

\boxed{y^2=\frac1{16}x-\frac1{128}},

\boxed{\alpha=\frac1{16},\quad\beta=-\frac1{128}}.

FP1 2025 Jan Q9

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程