题目

Problem

Figure 2 shows a sketch of the curve $C$ with equation

x = \frac{2}{3}\sin\left(3y + \frac{\pi}{4}\right)\qquad \frac{\pi}{12} < y < \frac{3\pi}{4}

The curve intersects the $y$ -axis at the points $A$ and $B$ as shown.

(a) Find the exact value of the $y$ coordinate of

point $A$
point $B$

(3)

(b) Show that

\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = \frac{1}{p - qx^2}

where $p$ and $q$ are integers to be found.

(4)

The normal to $C$ at $A$ and the tangent to $C$ at $B$ intersect at the point $D$ .

Using

the answer to part (b)
the sketch of curve $C$ in Figure 2

(4)

题目中文翻译

图 2 给出了曲线 $C$ 的草图，其方程为

x = \frac{2}{3}\sin\left(3y + \frac{\pi}{4}\right)\qquad \frac{\pi}{12} < y < \frac{3\pi}{4}

曲线与 $y$ 轴相交于点 $A$ 和 $B$ ，如图所示。

(a) 求下列各点的 $y$ 坐标的精确值：

点 $A$
点 $B$

(b) 证明：

\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = \frac{1}{p - qx^2}

其中 $p$ 和 $q$ 是待求的整数。

曲线 $C$ 在点 $A$ 处的法线与曲线 $C$ 在点 $B$ 处的切线相交于点 $D$ 。

利用：

(b) 的答案
图 2 中曲線 $C$ 的草圖

解答

(a)

The curve intersects the $y$ -axis when

x=0

\frac{2}{3}\sin\left(3y+\frac{\pi}{4}\right)=0

Hence

\sin\left(3y+\frac{\pi}{4}\right)=0

Given

\frac{\pi}{12}<y<\frac{3\pi}{4}

we have

\frac{\pi}{2}<3y+\frac{\pi}{4}<\frac{5\pi}{2}

In this interval,

3y+\frac{\pi}{4}=\pi

3y+\frac{\pi}{4}=2\pi

For the first value,

3y=\pi-\frac{\pi}{4}=\frac{3\pi}{4}

y=\frac{\pi}{4}

For the second value,

3y=2\pi-\frac{\pi}{4}=\frac{7\pi}{4}

y=\frac{7\pi}{12}

From the sketch, $A$ is the lower intersection and $B$ is the upper intersection.

Therefore

\boxed{y_A=\frac{\pi}{4}},\qquad \boxed{y_B=\frac{7\pi}{12}}

(b)

Differentiate

x=\frac{2}{3}\sin\left(3y+\frac{\pi}{4}\right)

with respect to $y$ :

\frac{\mathrm{d}x}{\mathrm{d}y} =\frac{2}{3}\cdot 3\cos\left(3y+\frac{\pi}{4}\right) =2\cos\left(3y+\frac{\pi}{4}\right)

Therefore

\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{1}{2\cos\left(3y+\frac{\pi}{4}\right)}

Squaring gives

\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 =\frac{1}{4\cos^2\left(3y+\frac{\pi}{4}\right)}

Now use

\cos^2 u=1-\sin^2 u

where

u=3y+\frac{\pi}{4}

From the curve equation,

x=\frac23\sin u

\sin u=\frac{3x}{2}

Hence

\begin{aligned} \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 &=\frac{1}{4\left(1-\sin^2u\right)} \\ &=\frac{1}{4\left(1-\left(\frac{3x}{2}\right)^2\right)} \\ &=\frac{1}{4-9x^2} \end{aligned}

Therefore

\boxed{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2=\frac{1}{4-9x^2}}

\boxed{p=4,\quad q=9}

(c)

At both $A$ and $B$ , we have $x=0$ .

Using part (b),

\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 =\frac{1}{4}

\frac{\mathrm{d}y}{\mathrm{d}x}=\pm\frac12

From the sketch, the tangent at $B$ has positive gradient, so its gradient is

\frac12

The normal at $A$ also has positive gradient. Since the tangent gradient at $A$ is negative, the normal gradient at $A$ is

2

The normal at $A\left(0,\frac{\pi}{4}\right)$ is

y-\frac{\pi}{4}=2x

The tangent at $B\left(0,\frac{7\pi}{12}\right)$ is

y-\frac{7\pi}{12}=\frac12x

At their intersection $D$ ,

2x+\frac{\pi}{4}=\frac12x+\frac{7\pi}{12}

\begin{aligned} \frac32x&=\frac{7\pi}{12}-\frac{\pi}{4} \\ \frac32x&=\frac{7\pi}{12}-\frac{3\pi}{12} \\ \frac32x&=\frac{\pi}{3} \end{aligned}

Therefore

x=\frac{2\pi}{9}

Hence the exact $x$ coordinate of $D$ is

\boxed{\frac{2\pi}{9}}

IAL 2025 May Q9

题目

解答

(a)

(b)

(c)