题目

Problem

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

3x^2+2y^2-4xy+8^x-11=0

The point $P$ has coordinates $(1,2)$ .

(a) Verify that $P$ lies on $C$ .

(1)

(b) Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .

(5)

The normal to $C$ at $P$ crosses the $x$ -axis at a point $Q$ .

(3)

题目中文翻译

图 2 给出了曲线 $C$ 的草图，其方程为

3x^2+2y^2-4xy+8^x-11=0

点 $P$ 的坐标为 $(1,2)$ 。

(a) 验证点 $P$ 在曲线 $C$ 上。

(b) 用 $x,y$ 表示 $\dfrac{dy}{dx}$ 。

曲线 $C$ 在点 $P$ 处的法线与 $x$ 轴交于点 $Q$ 。

解答

Substitute $x=1,\ y=2$ into the equation of $C$ :

3(1)^2+2(2)^2-4(1)(2)+8^1-11

This gives

3+8-8+8-11=0

So $P(1,2)$ lies on $C$ .

The equation of $C$ is

3x^2+2y^2-4xy+8^x-11=0

Differentiate implicitly with respect to $x$ .

Term by term,

\frac{d}{dx}(3x^2)=6x

and

\frac{d}{dx}(2y^2)=4y\frac{dy}{dx}

For $-4xy$ , use the product rule:

\frac{d}{dx}(-4xy) =-4\left(x\frac{dy}{dx}+y\right) =-4x\frac{dy}{dx}-4y

Also,

\frac{d}{dx}(8^x)=8^x\ln8

Therefore

6x+4y\frac{dy}{dx}-4x\frac{dy}{dx}-4y+8^x\ln8=0

Collect the terms involving $\dfrac{dy}{dx}$ :

(4y-4x)\frac{dy}{dx}+6x-4y+8^x\ln8=0

(4y-4x)\frac{dy}{dx} =4y-6x-8^x\ln8

Hence

\boxed{ \frac{dy}{dx} =\frac{4y-6x-8^x\ln8}{4y-4x} }

At $P(1,2)$ ,

\frac{dy}{dx} =\frac{4(2)-6(1)-8^1\ln8}{4(2)-4(1)}

\frac{dy}{dx} =\frac{8-6-8\ln8}{4} =\frac{2-8\ln8}{4}

This is the gradient of the tangent at $P$ .

Therefore the gradient of the normal is

-\frac{1}{\frac{2-8\ln8}{4}} =\frac{4}{8\ln8-2}

The equation of the normal is

y-2=\frac{4}{8\ln8-2}(x-1)

At $Q$ , the normal crosses the $x$ -axis, so

y=0

Substitute $y=0$ :

-2=\frac{4}{8\ln8-2}(x-1)

Thus

x-1=-\frac12(8\ln8-2)

x=1-\frac12(8\ln8-2)

Hence

x=2-4\ln8

Since

\ln8=\ln(2^3)=3\ln2

we get

x=2-12\ln2

Therefore the $x$ coordinate of $Q$ is

\boxed{2-12\ln2}