IS Math-2015-02

题目来源：Problem 2 日期：2024-08-11 题目主题：Math-Calculus-Curves on the $xy$-Plane

解题思路

本题主要考察了椭圆和双曲线的焦点性质，以及椭圆和双曲线族满足的微分方程。解题过程涉及以下关键步骤：

1. 利用椭圆和双曲线的几何定义推导焦点位置。
2. 构建和求解描述曲线族的微分方程。
3. 应用参数方程简化复杂的代数表达式。
4. 推导正交曲线族的方程。
5. 解析最终得到的曲线方程，确定其几何特征。

每个步骤都建立在前一个问题的基础上，形成了一个连贯的思路链，需要综合运用代数、微积分和几何知识。

Solution

Question 1

Ellipse

Given the ellipse:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0),$$

we are to show that the foci of this ellipse are $\left(\pm \sqrt{a^2-b^2},0\right)$.

Step 1: Geometric Definition of the Ellipse

The geometric definition of an ellipse states that for any point $\mathbf{P}=(x,y)$ on the ellipse, the sum of the distances to two fixed points, called the foci ${\mathbf{F}}_1$ and ${\mathbf{F}}_2$, is constant. Let the foci be at $(c,0)$ and $(-c,0)$. Thus, the condition is:

$$\sqrt{(x-c{)}^2+y^2}+\sqrt{(x+c{)}^2+y^2}=C.$$

Here, $C$ is the constant sum of the distances.

Step 2: Determine the Constant Using a Specific Point

Consider the special point $\mathbf{P}=(a,0)$ on the ellipse. Substituting into the distance equation:

$$\sqrt{(a-c{)}^2+0^2}+\sqrt{(a+c{)}^2+0^2}=C.$$

This simplifies to:

$$\sqrt{(a-c{)}^2}+\sqrt{(a+c{)}^2}=C.$$

Since both square roots are non-negative, we have:

$$|a-c|+|a+c|=C.$$

Since $a>0$ and $c$ must also be positive (as it represents the distance from the origin to the foci along the $x$-axis), the absolute values simplify to:

$$(a-c)+(a+c)=C.$$

This confirms that the constant sum is indeed $2a$. Therefore, we have validated that $2a$ is the sum of the distances from any point on the ellipse to the foci.

Recall that for an ellipse, the sum of the distances from any point $\mathbf{P}=(x,y)$ on the ellipse to the two fixed points (the foci) is constant and equal to $2a$. Let the foci be at $(c,0)$ and $(-c,0)$. The condition is:

$$\sqrt{(x-c{)}^2+y^2}+\sqrt{(x+c{)}^2+y^2}=2a.\text{(1)}$$

Step 3: Isolate and Square the Equation

We can start by isolating one of the square root terms in equation $(1)$:

$$\sqrt{(x-c{)}^2+y^2}=2a-\sqrt{(x+c{)}^2+y^2}.$$

Now, square both sides to eliminate the square root on the left side:

$$(x-c{)}^2+y^2={\left(2a-\sqrt{(x+c{)}^2+y^2}\right)}^2.$$

Expanding the square on the right side gives:

$$(x-c{)}^2+y^2=4a^2-4a\sqrt{(x+c{)}^2+y^2}+\left[(x+c{)}^2+y^2\right].$$

Now, let’s simplify by expanding $(x-c{)}^2$ and $(x+c{)}^2$:

$$x^2-2xc+c^2+y^2=4a^2-4a\sqrt{x^2+2xc+c^2+y^2}+x^2+2xc+c^2+y^2.$$

Next, cancel the common terms $x^2+c^2+y^2$ on both sides:

$$-2xc=4a^2-4a\sqrt{x^2+2xc+c^2+y^2}+2xc.$$

Bring the $4a\sqrt{x^2+2xc+c^2+y^2}$ term to the left side:

$$4a\sqrt{x^2+2xc+c^2+y^2}=4a^2+4xc.$$

Divide through by $4a$:

$$\sqrt{x^2+2xc+c^2+y^2}=a+\frac{xc}{a}.$$

Step 4: Square Again to Solve for $c$

Square both sides again to remove the square root:

$$x^2+2xc+c^2+y^2=a^2+\frac{2axc}{a}+\frac{x^2{c}^2}{a^2}.$$

Next, we substitute $y^2$ from the ellipse equation:

$$y^2=b^2\left(1-\frac{x^2}{a^2}\right).$$

Substitute this expression for $y^2$ into equation:

$$x^2+2xc+c^2+b^2\left(1-\frac{x^2}{a^2}\right)=a^2+\frac{2xc}{a}+\frac{x^2{c}^2}{a^2}.$$

Simplify the left-hand side:

$$x^2+2xc+c^2+b^2-\frac{b^2{x}^2}{a^2}=a^2+\frac{2xc}{a}+\frac{x^2{c}^2}{a^2}$$

Step 5: Combine and Simplify Terms

Group all terms involving $x^2$ on one side:

$$x^2\left(1-\frac{b^2}{a^2}\right)+2xc+c^2+b^2=a^2+\frac{2xc}{a}+\frac{x^2{c}^2}{a^2}.$$

Now, recall that $1-\frac{b^2}{a^2}=\frac{a^2-b^2}{a^2}$, so we can rewrite:

$$\frac{(a^2-b^2)x^2}{a^2}+2xc+c^2+b^2=a^2+\frac{2xc}{a}+\frac{x^2{c}^2}{a^2}.$$

Multiply through by $a^2$ to eliminate the fractions:

$$(a^2-b^2)x^2+2a^{2}xc+a^2{c}^2+a^2{b}^2=a^4+2a^{2}xc+x^2{c}^2.$$

Now, cancel $2a^{2}xc$ on both sides:

$$(a^2-b^2)x^2+a^2{c}^2+a^2{b}^2=a^4+x^2{c}^2.$$

Finally, collect all terms involving $x^2$:

$$(a^2-b^2-c^2)x^2=a^4-a^2{b}^2-a^2{c}^2.$$

For this to hold for all $x^2$, we must have:

$$a^2-b^2=c^2.$$

Thus, the foci are at $\left(\pm \sqrt{a^2-b^2},0\right)$, as required.

Hyperbola

Given the hyperbola:

$$\frac{x^2}{c^2}-\frac{y^2}{d^2}=1(c>d>0),$$

we are to show that the foci of this hyperbola are $\left(\pm \sqrt{c^2+d^2},0\right)$.

Step 1: Geometric Definition of the Hyperbola

The geometric definition of a hyperbola states that for any point $\mathbf{P}=(x,y)$ on the hyperbola, the absolute difference of the distances to two fixed points, called the foci ${\mathbf{F}}_1$ and ${\mathbf{F}}_2$, is constant. Let the foci be at $(f,0)$ and $(-f,0)$. Thus, the condition is:

$$\left|\sqrt{(x-f{)}^2+y^2}-\sqrt{(x+f{)}^2+y^2}\right|=C.$$

Here, $C$ is the constant difference of the distances.

Step 2: Determine the Constant Using a Specific Point

Consider the special point $\mathbf{P}=(c,0)$ on the hyperbola. Substituting into the distance equation:

$$\left|\sqrt{(c-f{)}^2+0^2}-\sqrt{(c+f{)}^2+0^2}\right|=C.$$

This simplifies to:

$$\left||c-f|-|c+f|\right|=C.$$

Since $f>0$, we have:

$$|(c-f)-(c+f)|=|-2f|=2f=C.$$

Thus, the constant difference is $2c$. Therefore, we have validated that $2c$ is the difference of the distances from any point on the hyperbola to the foci.

Now, using the specific point $(x,y)=(c,0)$, the constant $C=2c$, we rewrite the general equation for any point $(x,y)$ on the hyperbola:

$$\left|\sqrt{(x-f{)}^2+y^2}-\sqrt{(x+f{)}^2+y^2}\right|=2c.\text{(1)}$$

Step 3: Isolate and Square the Equation

Let’s isolate one of the square root terms in equation $(1)$:

$$\sqrt{(x-f{)}^2+y^2}=2c+\sqrt{(x+f{)}^2+y^2}.$$

Now, square both sides to eliminate the square root on the left-hand side:

$$(x-f{)}^2+y^2={\left(2c+\sqrt{(x+f{)}^2+y^2}\right)}^2.$$

Expanding the square on the right side gives:

$$(x-f{)}^2+y^2=4c^2+4c\sqrt{(x+f{)}^2+y^2}+\left[(x+f{)}^2+y^2\right].$$

Now, let’s simplify by expanding $(x-f{)}^2$ and $(x+f{)}^2$:

$$x^2-2xf+f^2+y^2=4c^2+4c\sqrt{x^2+2xf+f^2+y^2}+x^2+2xf+f^2+y^2.$$

Next, cancel the common terms $x^2+f^2+y^2$ on both sides:

$$-2xf=4c^2+4c\sqrt{x^2+2xf+f^2+y^2}+2xf.$$

Bring the $4c\sqrt{x^2+2xf+f^2+y^2}$ term to the left-hand side:

$$-4c\sqrt{x^2+2xf+f^2+y^2}=4c^2+4xf.$$

Divide through by $4c$:

$$-\sqrt{x^2+2xf+f^2+y^2}=c+\frac{xf}{c}.$$

Multiply by $-1$:

$$\sqrt{x^2+2xf+f^2+y^2}=-c-\frac{xf}{c}.$$

Step 4: Square Again to Solve for $f$

Square both sides again to remove the square root:

$$x^2+2xf+f^2+y^2=c^2+\frac{2cxf}{c}+\frac{x^2{f}^2}{c^2}.$$

Next, we substitute $y^2$ from the hyperbola equation:

$$y^2=d^2\left(\frac{x^2}{c^2}-1\right).$$

Substitute this expression for $y^2$ into the equation:

$$x^2+2xf+f^2+d^2\left(\frac{x^2}{c^2}-1\right)=c^2+\frac{2cxf}{c}+\frac{x^2{f}^2}{c^2}.$$

Simplify the left-hand side:

$$x^2+2xf+f^2+d^2\left(\frac{x^2}{c^2}\right)-d^2=c^2+\frac{2xf}{c}+\frac{x^2{f}^2}{c^2}.$$

Step 5: Combine and Simplify Terms

Group all terms involving $x^2$ on one side:

$$x^2\left(1+\frac{d^2}{c^2}\right)+2xf+f^2-d^2=c^2+\frac{2xf}{c}+\frac{x^2{f}^2}{c^2}.$$

Recall that $1+\frac{d^2}{c^2}=\frac{c^2+d^2}{c^2}$, so we can rewrite:

$$\frac{(c^2+d^2)x^2}{c^2}+2xf+f^2-d^2=c^2+\frac{2xf}{c}+\frac{x^2{f}^2}{c^2}.$$

Multiply through by $c^2$ to eliminate the fractions:

$$(c^2+d^2)x^2+2c^{2}xf+c^2{f}^2-c^2{d}^2=c^4+2c^{2}xf+x^2{f}^2.$$

Now, cancel $2c^{2}xf$ on both sides:

$$(c^2+d^2)x^2+c^2{f}^2-c^2{d}^2=c^4+x^2{f}^2.$$

Finally, collect all terms involving $x^2$:

$$(c^2+d^2-f^2)x^2=c^4-c^2{d}^2-c^2{f}^2.$$

For this to hold for all $x^2$, we must have:

$$c^2+d^2=f^2.$$

Thus, the foci are at $\left(\pm \sqrt{c^2+d^2},0\right)$, as required.

Question 2

The equation of an ellipse is:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{(Equation 1)}$$

where $a>b>0$, and the condition $a^2-b^2=u^2$ where $u>0$.

We need to show that any ellipse in the set $E_u$ satisfies the following differential equation:

$$xy(y^{\prime }{)}^2+\left(x^2-y^2-u^2\right)y^{\prime }-xy=0,$$

where $y^{\prime }=\frac{\mathrm{d}y}{\mathrm{d}x}$.

Step 1: Differentiate the ellipse equation

First, differentiate both sides of Equation 1 with respect to $x$:

$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)=\frac{\mathrm{d}}{\mathrm{d}x}(1)$$

This gives us:

$$\frac{2x}{a^2}+\frac{2y}{b^2}\cdot \frac{\mathrm{d}y}{\mathrm{d}x}=0$$

Simplify the above equation:

$$\frac{x}{a^2}+\frac{y}{b^2}{y}^{\prime }=0\text{(Equation 2)}$$

Step 2: Substitute $a^2=u^2+b^2$ into the ellipse equation

Using the condition $a^2=u^2+b^2$, substitute $a^2$ in Equation 1:

$$\frac{x^2}{u^2+b^2}+\frac{y^2}{b^2}=1\text{(Equation 3)}$$

Step 3: Express $b^2$ in terms of $x$, $y$, and $y^{\prime }$

From Equation 2:

$$\frac{x}{a^2}=-\frac{y}{b^2}{y}^{\prime }$$

Substitute $a^2=u^2+b^2$ into this equation:

$$\frac{x}{u^2+b^2}=-\frac{y}{b^2}{y}^{\prime }$$

Cross-multiply to solve for $b^2$:

$$b^2\cdot x=-y^{\prime }\cdot y\cdot (u^2+b^2)$$

Rearrange this equation to isolate $b^2$:

$$b^2\cdot (x+y\cdot y^{\prime })=-y^{\prime }\cdot y\cdot u^2$$

Finally, solve for $b^2$:

$$b^2=\frac{-y^{\prime }\cdot y\cdot u^2}{x+y\cdot y^{\prime }}$$

Step 4: Substitute $b^2$ into Equation 3

Substitute the expression for $b^2$ back into Equation 3:

$$\frac{x^2}{u^2+\frac{-y^{\prime }\cdot y\cdot u^2}{x+y\cdot y^{\prime }}}+\frac{y^2}{\frac{-y^{\prime }\cdot y\cdot u^2}{x+y\cdot y^{\prime }}}=1$$

Simplify the first term in the equation:

$$\frac{x^2\cdot (x+y\cdot y^{\prime })}{u^2(x+y\cdot y^{\prime })+y^{\prime }\cdot y\cdot u^2}$$

Simplify the equation to:

$$\frac{x^2}{u^2}\cdot \frac{(x+y\cdot y^{\prime })}{x}+\frac{y^2\cdot (x+y\cdot y^{\prime })}{-y^{\prime }\cdot y\cdot u^2}=1$$ $$(x+y\cdot y^{\prime })(x-\frac{y}{y^{\prime }})=u^2$$

Finally, expand and rearrange the terms to obtain the desired differential equation (***) in its standard form:

$$xy(y^{\prime }{)}^2+\left(x^2-y^2-u^2\right)y^{\prime }-xy=0$$

This completes the derivation, showing that any ellipse in the set $E_u$ satisfies the given differential equation.

Question 3

We can utilize the result from the ellipse problem by recognizing that the equations for ellipses and hyperbolas are structurally similar, with key differences in the signs.

Mapping the Ellipse to the Hyperbola

Given the ellipse equation:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

For the hyperbola, we have:

$$\frac{x^2}{c^2}-\frac{y^2}{d^2}=1$$

If we make the substitutions:

$$a^2\to c^2,b^2\to -d^2$$

This substitution directly transforms the ellipse’s equation into the hyperbola’s equation.

Utilizing the Conclusion from the Ellipse Problem

From the ellipse problem, we derived the differential equation:

$$xy(y^{\prime }{)}^2+\left(x^2-y^2-u^2\right)y^{\prime }-xy=0$$

By applying the same logic with the substitution $a^2=c^2$ and $b^2=-d^2$, we can directly use the result from the ellipse problem to conclude that any hyperbola in the set $H_u$, where $c^2+d^2=u^2$, will also satisfy the same differential equation:

$$xy(y^{\prime }{)}^2+\left(x^2-y^2-u^2\right)y^{\prime }-xy=0$$

Question 4: Finding the Differential Equation for $D_u$

We start with the problem of finding the differential equation that describes the set $D_u$ of curves that are perpendicular to any ellipse in the set $E_u$, where $E_u$ is defined as the set of ellipses given by:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

with the constraint $a^2-b^2=u^2$ for some positive constant $u$.

Step 1: Finding the Slope of the Tangent to the Ellipse

First, differentiate the equation of the ellipse implicitly with respect to $x$:

$$\frac{2x}{a^2}+\frac{2y}{b^2}\frac{\mathrm{d}y}{\mathrm{d}x}=0$$

This simplifies to:

$$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{b^2}{a^2}\cdot \frac{x}{y}$$

So, the slope of the tangent to the ellipse at any point $(x,y)$ is:

$$y_e^{\prime }=-\frac{b^{2}x}{a^{2}y}$$

Step 2: Finding the Slope of the Perpendicular Curves

For curves that are perpendicular to the ellipses in $E_u$, the slope $y^{\prime }$ of these curves must satisfy:

$$y^{\prime }\cdot y_e^{\prime }=-1$$

Since we know that this equation from (3) must hold for any ellipses in $E_u$:

$$xy(y_e^{\prime }{)}^2-(x^2-y^2-u^2)y_e^{\prime }-xy=0$$

We can simply substitute $y_e^{\prime }=-\frac{b^{2}x}{a^{2}y}$ into the equation above to find the differential equation for the set $D_u$：

$$xy(y^{\prime }{)}^2-(x^2-y^2-u^2)y^{\prime }-xy=0$$

Conclusion

The final differential equation for the set $D_u$ is:

$$\boxed{xy(y^{\prime }{)}^2-(x^2-y^2-u^2)y^{\prime }-xy=0}$$

This differential equation characterizes the family of curves perpendicular to the ellipses in $E_u$, excluding those that include a point where $y^{\prime }=0$ or where $x=0$.

Question 5: Solving the Differential Equation

Step 1: Use $\alpha$, $\beta$, $p$ to represent the Differential Equation

$\alpha {\beta }^{\prime }({\beta }^{\prime }+1)=\beta ({\beta }^{\prime }+1)+u^2{\beta }^{\prime }$

This simplifies to:

$$\alpha ({\beta }^{\prime 2}+{\beta }^{\prime })-u^2{\beta }^{\prime }=\beta +{\beta }^{\prime }\beta$$

Expanding and rearranging:

$$\alpha {\beta }^{\prime 2}+(\alpha -u^2-\beta ){\beta }^{\prime }-\beta =0$$

Now, following the same steps as in the earlier solution, differentiate both sides with respect to $\alpha$:

$$2\alpha {\beta }^{\prime }{\beta }^{\prime \prime }+(\alpha +2\alpha {\beta }^{\prime }-\beta -u^2){\beta }^{\prime \prime }=0$$

This equation simplifies to:

$$(\alpha +2\alpha {\beta }^{\prime }-\beta -u^2){\beta }^{\prime \prime }=0$$

Step 2: Solve the Differential Equation

There are two cases:

Case 1: ${\beta }^{\prime \prime }=0$

If ${\beta }^{\prime \prime }=0$, then ${\beta }^{\prime }$ is a constant:

$${\beta }^{\prime }=k$$

Integrating with respect to $\alpha$:

$$\beta =k\alpha +C_1$$

Case 2: $\alpha +2\alpha {\beta }^{\prime }-\beta -u^2=0$

Solving for ${\beta }^{\prime }$:

$$2\alpha {\beta }^{\prime }=\beta +u^2-\alpha$$

This simplifies to:

$${\beta }^{\prime }=\frac{\beta +u^2-\alpha }{2\alpha }$$

This is a first-order linear differential equation in $\beta (\alpha )$, which can be solved using the integrating factor method:

$$\beta =-u^2-\alpha +C_2\sqrt{\alpha }$$

Step 3: Relate $x$ and $y$ via the Conic Section Equation

Given the conditions:

$$\alpha =x^2\text{and}\beta =y^2$$

Substituting these into the solutions:

1. Linear Solution:

   $$y^2=k{x}^2+C_1$$

2. Non-linear Solutions:

   $$y^2=-x^2+2u\cdot x+C_2$$

   and

   $$y^2=-x^2-2u\cdot x+C_3$$

From the solutions, we have the following general forms:

1. Linear Solution:

   $$y^2=C_1{x}^2+C_2$$

2. Non-linear Solutions:

   $$y^2=-x^2+2u\cdot x+C_3$$ $$y^2=-x^2-2u\cdot x+C_4$$

Step 4: Completing the Square for Non-linear Solutions

Let’s complete the square for the non-linear solutions to express them as standard conic sections.

For $y^2=-x^2+2u\cdot x+C_3$

Start by rearranging terms involving $x$:

$$y^2=-(x^2-2ux)+C_3$$

Complete the square inside the parentheses:

$$y^2=-\left[(x^2-2ux+u^2)-u^2\right]+C_3$$

Simplify:

$$y^2=-\left[(x-u{)}^2-u^2\right]+C_3$$ $$y^2=-(x-u{)}^2+u^2+C_3$$

Now combine constants:

$$(x-u{)}^2+y^2=C_5$$

Where $C_5=u^2+C_3$.

For $y^2=-x^2-2u\cdot x+C_4$

Similarly, complete the square:

$$y^2=-(x^2+2ux)+C_4$$

Complete the square:

$$y^2=-\left[(x^2+2ux+u^2)-u^2\right]+C_4$$ $$y^2=-\left[(x+u{)}^2-u^2\right]+C_4$$

Simplify:

$$y^2=-(x+u{)}^2+u^2+C_4$$ $$(x+u{)}^2+y^2=C_6$$

Where $C_6=u^2+C_4$.

Step 5: Express the Linear Solution

For the linear solution:

$$C_1{x}^2-y^2=C_2$$

This is already in a standard form and doesn’t require further completion of the square.

Step 6: Final Conic Section Forms

We can now write the final conic section equations with all constants labeled as $C_i$:

1. Linear Solution:

$$C_1{x}^2-y^2=C_2$$

• This represents a parabola if $C_1=1$ and $C_2=0$.
• It represents a hyperbola if $C_1\ne 1$ or $C_2\ne 0$.

2. Non-linear Solution 1:

$$(x-u{)}^2+y^2=C_5$$

• The center is shifted to $(u,0)$.

3. Non-linear Solution 2:

   $$(x+u{)}^2+y^2=C_6$$
   • The center is shifted to $(-u,0)$.

In summary, the differential equations lead to conic sections, with the form of these conics (ellipse, hyperbola, or parabola) depending on the specific values of the constants $C_1,C_2,C_5,$ and $C_6$.

知识点

椭圆双曲线微分方程参数方程

难点思路

1. 焦点位置的推导：

   • 利用椭圆和双曲线的几何定义
   • 通过代数方法验证几何直观

2. 微分方程的构建与求解：

   • 从曲线方程出发，使用隐函数求导
   • 注意保留所有变量，避免过早代入具体值

3. 正交曲线族的推导：

   • 利用斜率的正交关系：$y^{\prime }\cdot y_{\text{original}}^{\prime }=-1$
   • 巧妙替换原方程中的 $y^{\prime }$ 为 $-\frac{1}{y^{\prime }}$

4. 参数方程的应用：

   • 引入参数简化复杂代数表达式
   • 利用参数方程揭示曲线的对称性和几何特征

5. 最终曲线方程的解析：

   • 使用完全平方法识别曲线类型
   • 分析不同参数值对曲线形状的影响

解题技巧和信息

1. 在处理椭圆和双曲线时，始终牢记其标准方程和几何定义的关系。
2. 解微分方程时，考虑使用分离变量法或积分因子法。
3. 在推导过程中，保持代数表达式的一般形式，避免过早代入特定值。
4. 利用对称性和几何直观简化计算和推理过程。
5. 在最后阶段，通过配方法将方程转化为标准形式，以便识别曲线类型。

重点词汇

• Ellipse: 椭圆
• Hyperbola: 双曲线
• Foci: 焦点
• Differential Equation: 微分方程
• Orthogonal Curves: 正交曲线
• Parametric Equation: 参数方程
• Implicit Differentiation: 隐函数求导

参考资料

1. Calculus by James Stewart, Chapter 10 (Parametric Equations and Polar Coordinates)
2. Differential Equations and Their Applications by Martin Braun, Chapter 1 (First-Order Differential Equations)
3. Thomas’ Calculus by George B. Thomas, Chapter 6 (Applications of Definite Integrals)
4. Linear Algebra and Its Applications by Gilbert Strang, Chapter 6 (Eigenvalues and Eigenvectors)
5. Schaum’s Outline of Differential Geometry by Martin M. Lipschutz, Chapter 2 (Curves in the Plane and in Space)