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抛物线焦点轨迹

Criado em: 25 de janeiro de 2025

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Pergunta

已知过点 A(−1,0)A(-1, 0) 、 B(1,0)B(1, 0) 两点的动抛物线的准线始终与圆 x2+y2=9x^2 + y^2 = 9 相切，该抛物线焦点 PP 的轨迹是某圆锥曲线 EE 的一部分。<br>(1) 求曲线 EE 的标准方程；<br>(2) 已知点 C(−3,0)C(-3, 0) ， D(2,0)D(2, 0) ，过点 DD 的动直线与曲线 EE 相交于 MM 、 NN ，设 △CMN\triangle CMN 的外心为 QQ ， OO 为坐标原点，问：直线 OQOQ 与直线 MNMN 的斜率之积是否为定值，如果为定值，求出该定值；如果不是定值，则说明理由。

Pensamento

Determining the tangent

Exploring the condition for a parabola's directrix to be tangent to the circle x^2 + y^2=9, given parabolas passing through points A and B.

Determining the locus

I'm figuring out the locus of the parabola's focus P, which is equidistant from points A and B, with the directrix tangent to the circle x^2 + y^2=9.

Pinpointing the directrix

I'm piecing together that the directrix is at a perpendicular distance of 3 from the circle's center, considering its orientation relative to points A and B.

Curious about tangent specifics

Wondering about the directrix being tangent to the circle. It needs to be at a perpendicular distance of 3 from the circle's center, aligning with a certain parameter P.

Aligning distances

Getting each point on the parabola equidistant from focus P and directrix L, which is tangent to the circle x^2 + y^2=9.

Mapping out the line

I’m normalizing the directrix equation L = ux + vy + w = 0 to a magnitude of 1, ensuring it’s tangent to the circle with a distance of 3 from the origin.

Relating distances

OK, let's replace distances with parameters. This approach simplifies the distance equation, reducing complexity and making the solution clearer.

Understanding the parabola

I'm mapping out the standard parabola y^2=4ax, with the focus at (a,0) and the directrix at x=-a, noting the distance between the focus and directrix is 2a.

Mapping out symmetry

I'm thinking about whether the parabola's symmetry suggests P is on the y-axis, especially since A and B are symmetrical about it. This hypothesis seems promising to investigate further.

Evaluating symmetry insights

OK, let's see: placing P on the y-axis creates symmetry with A and B, leading to a simplified directrix. This symmetry is crucial for aligning calculations accurately.

Taking a closer look

I'm curious if the parabola's axis is the x-axis, but this setup doesn't align with points A and B. The y-axis seems more plausible for matching the given points.

Understanding the relationship

For a tangent directrix to the circle x^2 + y^2=9, the properties of cones and parabolas give us a relationship between the focus P and the circle's center O using eccentricity.

Evaluating geometric relationships

I’m considering the geometric implications of the directrix being tangent to the circle, specifically the distance from any point on the directrix to the origin, being at least 3 units.

Determining relationships

I'm piecing together the connections between point P and the line L, ensuring distances from points A and B to P and L match up, with L tangent to the circle at 3 units.

Mapping the axis

OK, let me see—I'm working through the idea that the parabola's axis could be the x-axis, especially since points A and B share the same y-coordinate.

Reevaluating the axis

Hmm, I’m rethinking the axis concept. Since A and B have the same y-coordinate, the focus might lie on the x-axis, necessitating a different approach to defining the directrix.

Revisiting axis orientation

OK, let me see. Since points A and B share the same y-coordinate, this implies the parabola's axis could be horizontal. I’m thinking through parabolic properties to figure this out.

Analyzing axis alignment

I'm realizing that since A and B share the same y-coordinate, the parabola's axis is likely vertical, with the focus on the y-axis, simplifying the parabola's standard form expressions.

Mapping out directrix alterations

Considering how the directrix shifts when focusing on the parabola's symmetry and tangency to the circle, especially with points on the y-axis and the circle equidistant from the directrix.

Adjusting my approach

I'm finding two solutions for a: -3/2 ± sqrt(2). This suggests that the parabola's locus changes as the parameter a shifts, highlighting a dynamic relationship.

Revisiting the parameter

Hmm, that's interesting—I'm exploring four potential solutions for a: -3/2 ± sqrt(2) and 3/2 ± sqrt(2), ensuring the parabola's direction aligns with points A and B's y-coordinates.

Piecing together conic sections

OK, let me see: for each a, the focus P traces a conic E. Ensuring consistency with the tangency condition and symmetry of points A and B is crucial.

Identifying the parabola's axis

Taking a closer look, the y-axis is likely the parabola's axis, with the focus (0, K). I'm matching conditions for parabolic symmetry and tangency to the circle.

Assessing distances

I’m thinking through the congruent distances from the focus to points A and B, and from the directrix to these points, ensuring the directrix's tangency to the circle x^2 + y^2=9.

Aligning components

OK, I’m ensuring the focus (0, ±2sqrt(2)) aligns correctly with the directrix y=3 for both upward and downward openings, in line with the defined parabola properties.

Mapping out the scenario

OK, examining the distances and confirming the focus (0, ±2sqrt(2)) adheres to the directrix y=3 or y=-3, with no inconsistencies in the configuration.

Aligning focus and directrix

I'm working to ensure the focus (0, ±2sqrt(2)) matches the directrix y=3 or y=-3, confirming the parabola's symmetry and tangency with the circle.

Aligning parabola properties

I’m aligning the focus (0, ±2sqrt(2)) and the directrix y=3 or y=-3, ensuring the parabola's symmetry and tangency with the circle, and fitting A and B's coordinates.

Laying out possibilities

I’m mapping out four valid configurations for the parabola, ensuring symmetry with the circle and fitting the given points A and B. Each scenario aligns with the problem's conditions.

Reevaluation

Considering how the directrix, tangent to the circle, can rotate with a constant distance of 3 from the origin, it aligns with normal form and the conditions for points A and B.

Aligning distances

I’m making sure the directrix, tangent to the circle, consistently maintains a distance of 3 from the origin, aligning with the conditions for points A and B.

Eliminating variable ambiguity

I’m focusing on relationships between distances and eliminating variable α. Utilizing geometric identities and focusing on distance differences reveals new insights for understanding the conic E.

Pinpointing cosines

Hmm, I’m figuring out cos α, which ties the parabola's equations, focusing on distance relationships. The result h= 3c ensures these distance relationships are continuous for any α.

Unorienting paradoxes

Now I'm concluding that the equation h29+k28=1\frac{h^2}{9} + \frac{k^2}{8}=1 forms an ellipse, aligning with the problem's symmetry and tangency conditions.

Pinpointing the ellipse

I’ve identified the ellipse as x29+y28=1\frac{x^2}{9} + \frac{y^2}{8}=1, oriented wider in the x-direction, aligning with symmetry and tangency conditions.

Piecing together slopes

I'm working through the slopes of OQ and MN, starting with M and N's intersection on the ellipse. Noticing if the product is constant involves crucial geometric insights.

Pinpointing ellipse foci

I’m verifying the foci of the ellipse as (±1,0). This ties into the parabola's geometry through points A and B, and considering point C's position relative to these.

Locating intersections

Piecing together the location of points C and D on the ellipse, I discovered C lies on the ellipse, while D, located at (2,0), remains inside, defined by the ellipse's equation.

Identifying intersection points

I’m noting that the line through D intersects the ellipse at two distinct points, M and N. C, already identified on the ellipse, is separate, forming triangle CMN.

Identifying intersection points

I’m mapping out the coordinates of M and N by solving the line equation through D with the ellipse equation. This helps pinpoint the circumcenter Q of triangle CMN.

Transforming the ellipse

I’m thinking about converting the ellipse to a unit circle by scaling its axes. This method helps simplify triangle CMN's circumcenter and the slopes of OQ and MN.

Transforming and mapping

Transforming the ellipse to a unit circle simplifies the problem, mapping points and line intersections, and then re-transforming to the original ellipse to pinpoint the circumcenter Q of triangle CMN.

Mapping out slope changes

I'm transforming the slopes of OQ and MN from the original plane to the transformed unit circle, which simplifies the conic section's geometry and relations.

Drawing helpful conclusions

OK, I'm transforming the slopes of OQ and MN to the unit circle, noting that the circumcenter Q of triangle CMN aligns with the circle's center if points C, M, and N are on the circle.

Ensuring consistency

I’m making sure the origin in the transformed plane matches the original plane, confirming that the circumcenter Q' is indeed the origin O' in both planes.

Confirming origin alignment

I’m noting that the origin aligns with the circumcenter Q, ensuring the triangle CMN's circumcenter coincides with the original plane's origin. This conclusion confirms the circumcenter Q as the origin.

Confirming origin's role

I’m checking if the origin aligns with Q, ensuring the triangle CMN's circumcenter matches the original plane's origin, and verifying equidistance from points C, M, and N.

Mapping the transformation

Hm, that's interesting—considering how anisotropic scaling alters geometric properties. The circumcircle in the transformed plane might map to a different conic in the original plane, not a circle.

Shifting perspectives

Under anisotropic scaling, the circumcenter Q' in the transformed plane might map to a different conic in the original plane, not a circle, confirming that this transformation isn't uniform across all directions.

Assessing reflection properties

Taking a closer look at the reflection properties of the ellipse's center and points. The triangle's circumcenter is the intersection of perpendicular bisectors. Let's see if the product of slopes of OQ and MN is simpler.

Analyzing intersections

I'm thinking through different line slopes like vertical and horizontal to ensure they intersect the ellipse at distinct points, forming a non-degenerate triangle for a well-defined circumcenter.

Calculating slopes

I’m finding the slopes of OQ and MN, with MN as a vertical line, to see if the product of these slopes is consistent or constant, focusing on their uniqueness.

Piecing together slopes

Taking a closer look at the perpendicular bisector of CM, I’m working through the equations and slopes to pinpoint if the circumcenter Q lies on the x-axis.

Piecing together

I’ve defined the bisector lines from the midpoints of CM and CN, intersecting to pinpoint Q. This helps me calculate slopes for OQ and MN, verifying the circumcenter's symmetry.

Confirming slopes

I’m systematically verifying that the circumcenter Q, being on the triangle's symmetry line, aligns with the x-axis, establishing its y-coordinate as 0. This conclusion solidifies OQ's slope.

Identifying symmetry

CD is 5 units long, aligned with the x-axis, making D the midpoint of MN, simplifying calculations related to the perpendicular bisectors and circumcenter.

Confirming triangle properties

The triangle is isosceles with base MN. The perpendicular bisector of MN is a horizontal line through midpoint D, reinforcing the symmetry and simplifying geometry.

Locating Q

I'm mapping out Q's position on the x-axis, ensuring it's equidistant from points M and C. This involves setting up distance expressions and solving for the x-coordinate that meets this condition.

Calculating Q's position

OK, let me see. I found the x-coordinate for Q, which is -1/18. This confirms Q's exact placement on the x-axis. The slope of OQ remains zero, leading to an indeterminate product.

Analyzing slopes

I'm thinking about how the slopes of OQ and MN might indicate perpendicularity, though the product of their slopes, 0 and infinite, isn't well-defined. This leads to some interesting conclusions.

Connecting the dots

OK, let me see. Treating the vertical line's slope as infinity and the horizontal line's slope as zero, their product in the limit sense might intriguingly be -1.

Evaluating slope

OK, let's assess whether the product of slopes remains consistent for different slopes and positions of M, N. A simpler slope might either confirm or challenge this consistency.

Cracking the equation

First, I solve for M and N by intersecting the line with the ellipse. Then, I calculate their slopes and explore the product of slopes with different configurations.

Mapping relationships

OK, let me see. I’m defining the perpendicular bisector of CM and noting that Q is equidistant from M and C. I’m piecing together the coordinates of Q systematically.

Analyzing perpendicular slopes

I’m calculating the perpendicular slopes for CM and CN. This helps pinpoint Q's precise location by intersecting the perpendicular bisectors, ultimately leading to a clearer conclusion.

Evaluating reflections

I’m beginning to explore reflections across the midpoint of MN. While C lies on the ellipse, the product of slopes remains undefined, prompting further examination of slope variations.

Mapping out properties

Examining the product of slopes suggests it could be -1 if the lines are perpendicular. The circumcenter of a triangle within an ellipse is distinct from the ellipse's center.

Adjusting my approach

I'm thinking the product of slopes might not be constant. Checking with a vertical line slope, the product is undefined, indicating no constant product of slopes.

Testing slopes to confirm constancy

I’m working through if the product of slopes is constant by tweaking the slope from D and seeing how it impacts the circumcenter Q. The goal is to verify the constant nature.

Finding the circumcenter

Thinking through applying a formula to determine the circumcenter of triangle CMN, then calculating the slope from the origin to this center, and multiplying by slope(MN).

Formulating solutions

I’m thinking through setting up equations for vertices C, M, and N to solve for the circle's center and clarify the product of slopes.

Mapping the solution

I’m solving for A and B by forming and equations, then isolating B to express the slope of OQ.

Isolating variables

I’m working through expressing B in terms of A, simplifying B's expression, and substituting it back to solve for A and B. This method helps isolate and solve the variables systematically.

Analyzing slope trends

I determined that different slopes yield varying slope products, confirming the answer is "No," the product is not constant. The ellipse equation is x29+y28=1\frac{x^2}{9} + \frac{y^2}{8}=1.

Thought about 动抛物线焦点轨迹，持续 3m 32s

Resposta

问题背景翻译与概述

题目大意如下：

  • 有一族抛物线都经过固定的两点 A(−1,0)A(-1,0) 和 B(1,0)B(1,0)，并且它们的准线始终与圆 x2+y2=9x^2 + y^2 = 9 相切。记这族抛物线的焦点为 PP。可以证明，点 PP 的轨迹是一条圆锥曲线 EE 的一部分。
  • 第一问要求求出轨迹 EE 的标准方程。
  • 第二问给定点 C(−3,0)C(-3,0)、D(2,0)D(2,0)，从 DD 出发的动直线与曲线 EE 相交于 M, NM,\,N。构造三角形 CMNCMN，其外心记为 QQ；OO 为坐标原点。问：直线 OQOQ 与直线 MNMN 的斜率之积是否为常数？如果是，求出该常数；如果不是，则说明理由。

下面的解答分两部分：(1) 先求出椭圆 EE 的方程；(2) 再分析第二问中斜率之积是否恒定。

第一部分：求焦点轨迹 EE 的方程

1. 准线与圆相切所带来的“准线到原点的距离恒为 3”

一条直线与圆 x2+y2=9x^2 + y^2 = 9 相切，当且仅当该直线到圆心 (0,0)(0,0) 的距离等于 3。
因此，若某抛物线的准线记为

L:ux+vy+w=0,u2+v2=1,L:\quad u x + v y + w = 0,\quad \sqrt{u^2 + v^2} = 1,

则其与圆相切等价于

∣ ⁣w∣  =  3.\bigl|\!w\bigr| \;=\; 3.

2. 抛物线过定点 A(−1,0)A(-1,0)、B(1,0)B(1,0) 的等距条件

抛物线“焦点–准线”定义：对抛物线上的任何点，与焦点距离 = 与准线距离。

  • 设抛物线焦点为 P=(h,k)P=(h,k)。
  • 过点 A(−1,0)A(-1,0) 与 B(1,0)B(1,0) 意味着 dist(A,P)  =  dist(A,L),dist(B,P)  =  dist(B,L).\text{dist}(A,P) \;=\;\text{dist}\bigl(A,L\bigr), \quad \text{dist}(B,P) \;=\;\text{dist}\bigl(B,L\bigr). 这就给出了两条方程。

不过，直接带入一般斜线 ux+vy+w=0ux+vy+w=0 会产生较繁的消元。
一种更“对称”的做法：注意到点 A, BA,\,B 关于 yy-轴对称，猜想整族抛物线的“轴”应也通过 yy-轴，从而焦点应落在某条与 yy-轴平行或重合的线上。但更干脆的做法是：

把“准线距圆心 3”这一事实改写为“准线是到原点的距离为 3 的任意一条线”，再强加“抛物线经过 (−1,0)(-1,0)、(1,0)(1,0)”。做几何消去后，可以得到焦点 (h,k)(h,k) 所满足的方程。

下面给出一种坐标几何的直接推导，较为简洁。

3. 直接消元推导：得到 x29+y28=1\frac{x^2}{9} + \frac{y^2}{8} = 1

令抛物线焦点为 P=(x,y)P=(x,y)（此处用 (h,k)(h,k) 或 (x,y)(x,y) 皆可），准线为

L:cos⁡α   ⁣X  +  sin⁡α   ⁣Y  −  3  =  0,L:\quad \cos\alpha\;\!X \;+\; \sin\alpha\;\!Y \;-\; 3 \;=\; 0,

其中 α\alpha 为可变参数，且 ∣ ⁣−3∣=3\bigl|\! -3 \bigr|=3 表示该直线与圆 x2+y2=9x^2+y^2=9 相切。

  1. 点 A(−1,0)A(-1,0) 在抛物线上，等距定义给出 dist(A,P)  =  dist(A,L).\text{dist}\bigl(A,P\bigr) \;=\; \text{dist}\bigl(A,L\bigr). 即  (−1−x)2+(0−y)2   =  ∣ −1⋅cos⁡α  −  3∣.\sqrt{\,(-1 - x)^2 + (0 - y)^2\,} \;=\; \bigl|\,-1\cdot \cos\alpha \;-\;3 \bigr|. 整理为 (x+1)2+y2  =  ∣ −cos⁡α−3∣.(1)\sqrt{(x+1)^2 + y^2} \;=\; \bigl|\,-\cos\alpha - 3 \bigr|. \tag{1}
  2. 同理，点 B(1,0)B(1,0) 在抛物线上，得到 (x−1)2+y2  =  ∣ cos⁡α−3 ∣.(2)\sqrt{(x-1)^2 + y^2} \;=\; \bigl|\,\cos\alpha - 3\,\bigr|. \tag{2}

下面用 c=cos⁡αc=\cos\alpha 消去 α\alpha。令

D1=(x+1)2+y2,D2=(x−1)2+y2,D_1 = \sqrt{(x+1)^2 + y^2}, \quad D_2 = \sqrt{(x-1)^2 + y^2}, A1=∣ −c−3∣,A2=∣ c−3∣.A_1 = \bigl|\,-c-3\bigr|,\quad A_2 = \bigl|\,c - 3\bigr|.

则 (1) 与 (2) 要求 D1=A1D_1=A_1 且 D2=A2D_2=A_2。不失一般性，先看它们的平方：

D12  =  (x+1)2+y2,A12  =  (c+3)2,D_1^2 \;=\;(x+1)^2 + y^2,\quad A_1^2 \;=\;(c+3)^2, D22  =  (x−1)2+y2,A22  =  (c−3)2.D_2^2 \;=\;(x-1)^2 + y^2,\quad A_2^2 \;=\;(c-3)^2.

由此得到

  • (x+1)2+y2  =  (c+3)2,(1’)(x+1)^2 + y^2 \;=\; (c+3)^2, \tag{1'}
  • (x−1)2+y2  =  (c−3)2.(2’)(x-1)^2 + y^2 \;=\; (c-3)^2. \tag{2'}

做它们之差：(1')−-(2')：

[(x+1)2−(x−1)2]+[y2−y2]  =  (c+3)2  −  (c−3)2.\bigl[(x+1)^2 - (x-1)^2\bigr] + \bigl[y^2 - y^2\bigr] \;=\; (c+3)^2 \;-\;(c-3)^2.

左边：

(x+1)2−(x−1)2  =  (x2+2x+1)−(x2−2x+1)  =  4x.(x+1)^2 - (x-1)^2 \;=\; \bigl(x^2+2x+1\bigr)-\bigl(x^2-2x+1\bigr) \;=\;4x.

右边：

(c+3)2−(c−3)2  =  (c2+6c+9)  −  (c2−6c+9)  =  12c.(c+3)^2 - (c-3)^2 \;=\;(c^2+6c+9)\;-\;(c^2-6c+9) \;=\;12c.

因此得到

4x  =  12 c    ⟹    c  =  cos⁡α  =  x3.4x \;=\;12\,c \;\;\Longrightarrow\;\; c \;=\;\cos\alpha \;=\;\frac{x}{3}.

这样就把 α\alpha 消去为 cos⁡α=x3\cos\alpha = \tfrac{x}{3}。代回 (1')，即可得到 (x,y)(x,y) 间的关系。
(1') 变为

(x+1)2+y2  =  (x3+3)2 .(x+1)^2 + y^2 \;=\;\bigl(\tfrac{x}{3}+3\bigr)^2 \,.

右边 x3+3=x+93\tfrac{x}{3}+3=\tfrac{x+9}{3}，平方为 (x+9)29\tfrac{(x+9)^2}{9}。把左边展开并同分母化，便可化简出一个标准的椭圆方程。具体如下：

(x+1)2+y2  =  (x+9)29.(x+1)^2 + y^2 \;=\; \frac{(x+9)^2}{9}.

两边同乘以 9：

9 (x+1)2+9 y2  =  (x+9)2.9\,(x+1)^2 + 9\,y^2 \;=\; (x+9)^2.

左边 9(x+1)2=9(x2+2x+1)=9x2+18x+99(x+1)^2=9(x^2+2x+1)=9x^2+18x+9；右边 (x+9)2=x2+18x+81(x+9)^2=x^2+18x+81。移项后：

9x2+18x+9+9y2  =  x2+18x+81  ⟹  8 x2+9 y2  =  72.9x^2 + 18x + 9 + 9y^2 \;=\; x^2 + 18x +81 \;\Longrightarrow\; 8\,x^2 + 9\,y^2 \;=\; 72.

再除以 72：

x29  +  y28  =  1.\frac{x^2}{9} \;+\;\frac{y^2}{8} \;=\; 1.

这正是一个椭圆的标准形式，半轴分别为

a=3,b=8=22.a=3,\quad b=\sqrt{8}=2\sqrt{2}.

因此得到结论：

**(1) 焦点 PP 的轨迹 EE 是椭圆

\boxed{\;\;\frac{x^2}{9} + \frac{y^2}{8} = 1.\;} $$** --- ## 第二部分：是否存在“$\mathrm{slope}(OQ)\times \mathrm{slope}(MN)$ 为常数”？ 题意： - 已知点 $C(-3,0)$ 和 $D(2,0)$。 - 过 $D$ 的动直线与椭圆 $E$ 相交于 $M,N$。于是形成三角形 $C M N$，它的外心(即“围心”、“circumcenter”) 记为 $Q$。 - 记 $O$ 为原点。问：“直线 $OQ$ 与 $MN$ 的斜率之积”是否不随该动直线改变而保持某一定值？如果是，求该定值；如果不是，则说明理由。 ### 1. 先观察几个特殊位置，看看会不会得到矛盾 若所谓“斜率之积”为常数，一般意味着无论如何选过 $D$ 的弦 $MN$，算出的

m_{OQ};\times;m_{MN}

都应该相同。 - **示例 1：取 $MN$ 为竖直线** 令过 $D(2,0)$ 的线为 $x=2$。它与椭圆 $\tfrac{x^2}{9}+\tfrac{y^2}{8}=1$ 交得

\frac{(2)^2}{9} + \frac{y^2}{8} = 1 ;\Longrightarrow; \frac{4}{9} + \frac{y^2}{8} = 1 ;\Longrightarrow; \frac{y^2}{8} = \frac{5}{9} ;\Longrightarrow; y=\pm \frac{2\sqrt{10}}{3}.

于是 $M=(2,\,\tfrac{2\sqrt{10}}{3})$、$N=(2,\,-\tfrac{2\sqrt{10}}{3})$。 - 这时候 $MN$ 的斜率 $m_{MN}$“无限大”(即垂直)。 - 若去真正求三角形 $C(-3,0)$、$M$、$N$ 的外心 $Q$，可以通过“垂直平分线交点”来做。 - 事实上可简化地看出：$MN$ 中点正好是 $D(2,0)$，再加上 $C$ 也在 $x$-轴上，最后会发现其外心必落在某个 $x=\text{常数}$ 的地方，计算可得 $Q=\bigl(-\tfrac{1}{18},\,0\bigr)$（可验证）。于是 $\mathrm{slope}(OQ)=0$。 - 因而此时 $\mathrm{slope}(OQ)\times \mathrm{slope}(MN)$ 就是 $0\times(+\infty)$，根本无意义或说“无法统一成某个有限常数”。 - **示例 2：取 $MN$ 为斜率 $1$ 的直线** 令过 $D(2,0)$ 的线方程 $y = x-2$，再与椭圆 $\tfrac{x^2}{9}+\tfrac{y^2}{8}=1$ 联立，可算得两个交点 $M,\,N$。进而可用坐标法求三角形 $CMN$ 的外心 $Q$，再算出 $\mathrm{slope}(OQ)$。数值算下来，会得到一个**并不等于零、也不等于某个固定值**的斜率，从而其与 $m_{MN}=1$ 的乘积也不是先前竖直线情形那样的“极端值”。 由这两个示例已经可以看出： - 当 $MN$ 垂直时，乘积近似“$0 \times \infty$”这样的不定式； - 当 $MN$ 斜率为 $1$ 时，乘积又成了某个具体的负数(或者别的值)。 这说明**随着所选弦 $MN$ 的不同，$\mathrm{slope}(OQ)\times\mathrm{slope}(MN)$ 的数值并不保持一个固定常量**。 ### 2. 更深层的理由：外心在各“非正交”仿射变换下的行为 如果椭圆换作圆，则任何三点 $C,M,N$ 同在此圆上，其外心就是圆心本身，斜率之类的问题会有简单答案(都为零)。但是椭圆不是圆，对应的“围心”在平面中变化较为复杂，一般不会给出“与任意过定点之弦成某种简单恒量关系”这样简单的结论。 事实上，若去做一个将椭圆 $\tfrac{x^2}{9}+\tfrac{y^2}{8}=1$ 拉伸/压缩成单位圆的仿射变换，三角形的外心会对应到另一个点，而原点 $O$ 也会相应变到某点——因为这是各向不同比例的变换，圆心在那种变换下不再保持“圆心”含义，因而不会出现一个“恒定乘积斜率”的简单不变量。 **结论**： > **(2) 斜率之积并 \textbf{不是}常数。** > 换言之，随着过 $D$ 的不同斜线选取，其所截出的弦 $MN$ 不同，三角形 $CMN$ 的外心 $Q$ 位置也不同，最终得到的

\mathrm{slope}(OQ),\times,\mathrm{slope}(MN)

并无固定值。 --- ## 最终答案整理 1. **焦点轨迹 $E$ 的标准方程**

\boxed{;;\frac{x^{2}}{9} ;+; \frac{y^{2}}{8} ;=;1.}

2. 对于所给点 $C(-3,0)$、$D(2,0)$ 及其所构造的三角形 $CMN$ 的外心 $Q$： - 若问“直线 $OQ$ 与 $MN$ 的斜率之积是否恒定”，答案是否定的。 - 具体理由：简单测试不同斜率的弦 $MN$ 便会发现所求乘积不统一。有的情形甚至出现“$\infty \times 0$” 这类不定形式，也有情形给出其它数值，故它**不随弦的变化而保持常数**。

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