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已知过点 $A(-1, 0)$ 、 $B(1, 0)$ 两点的动抛物线的准线始终与圆 $x^2 + ...

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प्रश्न

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。

विचार

Analyzing the parabolic family

Given the two points A(-1, 0) and B(1, 0) on the parabola, the focus of the parabola must satisfy the condition that the directrix is tangent to the circle x^2 + y^2 = 9. This suggests a family of parabolas. To solve for the locus of the focus, we need to find the equation for the conic that contains these focus points.

For part (2), the geometry of the moving line intersecting the conic needs further exploration, aimed at understanding the relationship between the slopes.

Exploring the parabolic setup

The parabola has its directrix tangent to the circle x^2 + y^2 = 9, and the focus P is variable. For any parabola, the points (x, y) are equidistant to the focus and directrix. We're given that the parabola passes through points A(-1, 0) and B(1, 0), and the directrix touches the circle. To find the locus of the focus, we define the equations based on these geometric properties, with focus coordinates (p, q) and directrix parameters. Using this, I can start deriving conditions leveraging the parabola's focal property.

Refining the parabola's focus

The directrix has equation x cosθ + y sinθ = 3, and the distance from the center to the line gives the condition |c| = 3. The focus must lie on the line perpendicular to the directrix, and the distance from focus to directrix is related to the position of the vertex. I'm using the fact that points A(-1, 0) and B(1, 0) are on the parabola, so their distances from the focus must equal their distances to the directrix. There seems to be more geometric insight coming together here.

Solving focus locus and slope product

We've derived that the locus of the focus P is an ellipse given by the equation x^2/9 + y^2/8 = 1. For the second part, we are given point C(-3,0) and point D(2,0), and we need to analyze the product of slopes between the lines OQ (from the origin to the circumcenter Q of triangle CMN) and MN. This requires geometric insight into the ellipse, the moving line through D, and their intersections. We can explore the relationship step-by-step using analytical methods and the harmonic properties of the conic.

Analyzing the geometry of triangle CMN

We're analyzing the geometry of the circumcenter Q of triangle CMN, where C = (-3, 0) is fixed, and M and N are the intersections of the moving line L with ellipse E. Computing the slopes of the lines OQ (from the origin to Q) and MN requires geometric reasoning, leveraging the properties of perpendicular bisectors and the circumcenter. I'm building equations based on distances and slopes, but at this stage, I need to progress carefully with solving for Q's coordinates and ensuring any symmetries are consistently applied.

Geometric considerations for circumcenter Q

I’m working through the problem of finding the circumcenter Q of triangle CMN formed by the intersection points of the moving line with the ellipse. The approach involves solving for the relationship between the coordinates of Q, using properties like perpendicular bisectors and midpoint formulae. My goal is to eventually compute the product of slopes of OQ and MN, and verify if this product is constant. The solution involves several algebraic steps, each relating to the geometry of circles and lines. It's a dense but coherent process!

Finding the circumcenter Q

To find Q, I’m using the property that it lies on the perpendicular bisector of MN. From here, I’m trying to compute t, which positions Q relative to other points. My approach involves equations based on the distances from Q to C and M, incorporating ellipse and line intersection information. I’m calculating various distances such as the distance from Q to the chord MN. After some simplification, I express Q’s coordinates in terms of t and use known properties to generate a system of equations to solve for Q's position.

推理关于动抛物线的焦点轨迹，持续 1m 8s

उत्तर

下面给出本题的详细解法。

（1）求焦点轨迹 EE 的标准方程

设该抛物线的焦点为 P(p,q)P(p,q) ，准线为

L:xcos⁡θ+ysin⁡θ=c.L:\quad x\cos\theta+y\sin\theta= c.

由题意，准线始终与圆

x2+y2=9x^2+y^2=9

相切，因此准线到原点的距离恒为 3，即

∣c∣=3.|c|=3.

（注意：写成 xcos⁡θ+ysin⁡θ=3x\cos\theta+y\sin\theta=3 或 =−3=-3 均可；下面证明过程不因选取正负而改变，故下面我们取

L:xcos⁡θ+ysin⁡θ=3.L:\quad x\cos\theta+y\sin\theta=3.

抛物线的定义是：平面上到焦点的距离等于到准线的距离。因此，任一点 X(x,y)X(x,y) 在抛物线上满足

(x−p)2+(y−q)2=∣xcos⁡θ+ysin⁡θ−3∣1.\sqrt{(x-p)^2+(y-q)^2}=\frac{|x\cos\theta+y\sin\theta-3|}{1}.

题中给出抛物线必过定点 A(−1,0)A(-1,0) 和 B(1,0)B(1,0)，故有

(p+1)2+q2=∣−cos⁡θ−3∣,(p−1)2+q2=∣cos⁡θ−3∣.\sqrt{(p+1)^2+q^2}=\bigl| -\cos\theta-3\bigr|,\quad \sqrt{(p-1)^2+q^2}=\bigl| \cos\theta-3\bigr|.

为消去 θ\theta 我们取这两式平方后相减：

(p+1)2+q2−(p−1)2−q2=(cos⁡θ+3)2−(3−cos⁡θ)2[(p+1)2−(p−1)2]=[(cos⁡θ+3)2−(3−cos⁡θ)2].\begin{array}{rcl} (p+1)^2+q^2-(p-1)^2-q^2&=&(\cos\theta+3)^2-(3-\cos\theta)^2\\[1mm] \Bigl[(p+1)^2-(p-1)^2\Bigr] &=&\Bigl[(\cos\theta+3)^2-(3-\cos\theta)^2\Bigr]. \end{array}

计算左边：

(p+1)2−(p−1)2=(p2+2p+1)−(p2−2p+1)=4p.(p+1)^2-(p-1)^2=(p^2+2p+1)-(p^2-2p+1)=4p.

右边利用完全平方公式：

(cos⁡θ+3)2−(3−cos⁡θ)2=[cos⁡2θ+6cos⁡θ+9]−[cos⁡2θ−6cos⁡θ+9]=12cos⁡θ.(\cos\theta+3)^2-(3-\cos\theta)^2 =\Bigl[\cos^2\theta+6\cos\theta+9\Bigr]-\Bigl[\cos^2\theta-6\cos\theta+9\Bigr] =12\cos\theta.

故得

4p=12cos⁡θ⟹p=3cos⁡θ.4p=12\cos\theta\quad\Longrightarrow\quad p=3\cos\theta.

接下来将 p=3cos⁡θp=3\cos\theta 代入任一一式中（例如选用 B(1,0)B(1,0) 得）

(p−1)2+q2=(cos⁡θ−3)2.(p-1)^2+q^2=(\cos\theta-3)^2.

将 p=3cos⁡θp=3\cos\theta 写入：

(3cos⁡θ−1)2+q2=(cos⁡θ−3)2.(3\cos\theta-1)^2+q^2=(\cos\theta-3)^2.

不必再解 θ\theta ，因为我们只需要将 cos⁡θ\cos\theta用 pp 表示，即 cos⁡θ=p3\cos\theta=\frac{p}{3}。故有

(p−1)2+q2=(p3−3)2=(p−93)2=(p−9)29.\left(p-1\right)^2+q^2=\Bigl(\frac{p}{3}-3\Bigr)^2 =\left(\frac{p-9}{3}\right)^2=\frac{(p-9)^2}{9}.

把两边都乘以 9：

9[(p−1)2+q2]=(p−9)2.9\Bigl[(p-1)^2+q^2\Bigr]=(p-9)^2.

展开计算：

9[p2−2p+1+q2]=p2−18p+81.9\Bigl[p^2-2p+1+q^2\Bigr]=p^2-18p+81.

即

9p2−18p+9+9q2=p2−18p+81.9p^2-18p+9+9q^2=p^2-18p+81.

消去两边相同的 −18p-18p，整理得

9p2+9q2+9=p2+81.9p^2+9q^2+9=p^2+81.

移项整理：

8p2+9q2=72.8p^2+9q^2=72.

将其写成标准形式：

p29+q28=1.\frac{p^2}{9}+\frac{q^2}{8}=1.

记 x=p,  y=qx=p,\;y=q，则焦点 PP 的轨迹为

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

（2）证明直线 OQOQ 与 MNMN 斜率乘积为常数

题中已知定点 C(−3,0)C(-3,0) 和 D(2,0)D(2,0)。取过 DD 的动直线

L:y=k(x−2)L:\quad y=k(x-2)

与曲线 E: x29+y28=1E:\,\frac{x^2}{9}+\frac{y^2}{8}=1 相交于 MM 和 NN（当 k≠0k\ne0 时这两点不重），构成三角形 CMNCMN；设其外心为 QQ ，原点为 OO 。

下面证明：直线 MNMN 的斜率为 kk ，而经过 O(0,0)O(0,0) 与 QQ 的连线斜率恰为 −5 k\displaystyle -\frac{5}{\,k}，故二者乘积为

k⋅(−5k)=−5,k\cdot\Bigl(-\frac{5}{k}\Bigr)=-5,

与 kk 无关，是常值。

下面给出一种求 QQ 坐标的计算方法（注意：本法仅对非退化情形，即 k≠0k\ne0 时成立）。

（2.1）求外接圆的圆心 QQ

设三角形 CMNCMN 的外接圆（即过 C,M,NC, M, N 的圆）方程为

x2+y2+2Gx+2Fy+H=0.x^2+y^2+2Gx+2Fy+H=0.

其圆心为 Q=(−G,−F)Q=(-G,-F)．又因为该圆过 C(−3,0)C(-3,0)，故代入有

(−3)2+02+2G(−3)+H=0⟹9−6G+H=0,(-3)^2+0^2+2G(-3)+H=0\quad\Longrightarrow\quad 9-6G+H=0,

即

H=6G−9.H=6G-9.

另一方面，由于 MM 和 NN 均在直线

L:y=k(x−2)L:\quad y=k(x-2)

上，将 y=k(x−2)y=k(x-2) 代入圆的方程，得

x2+k2(x−2)2+2Gx+2Fk(x−2)+6G−9=0.x^2+k^2(x-2)^2+2Gx+2Fk(x-2)+6G-9=0.

整理后可写成关于 xx 的二次式：

(1+k2)x2+(2G+2Fk−4k2)x+(4k2−4Fk+6G−9)=0.\bigl(1+k^2\bigr)x^2+\Bigl(2G+2Fk-4k^2\Bigr)x+\Bigl(4k^2-4Fk+6G-9\Bigr)=0.

而 M,NM, N 又是曲线

E:x29+y28=1E:\quad \frac{x^2}{9}+\frac{y^2}{8}=1

与直线 LL 的交点。将 y=k(x−2)y=k(x-2) 代入 EE ，乘以 72 得

(8+9k2)x2−36k2x+(36k2−72)=0.(8+9k^2)x^2-36k^2 x+(36k^2-72)=0.

因为二者根集相同（系数只相差一个比例因子 λ\lambda ），所以存在 λ\lambda 使得

{1+k2=λ(8+9k2),2G+2Fk−4k2=λ(−36k2),4k2−4Fk+6G−9=λ(36k2−72).\begin{cases} 1+k^2=\lambda(8+9k^2),\\[1mm] 2G+2Fk-4k^2=\lambda(-36k^2),\\[1mm] 4k^2-4Fk+6G-9=\lambda(36k^2-72). \end{cases}

从第一式得

λ=1+k28+9k2.\lambda=\frac{1+k^2}{8+9k^2}.

而第二式则可写为

2G+2Fk=4k2−36λk2,2G+2Fk=4k^2-36\lambda k^2,

即

G+Fk=2k2(1−9λ).G+Fk=2k^2\Bigl(1-9\lambda\Bigr).

注意，由于

1−9λ=1−91+k28+9k2=8+9k2−9−9k28+9k2=−18+9k2,1-9\lambda=1-9\frac{1+k^2}{8+9k^2} =\frac{8+9k^2-9-9k^2}{8+9k^2} =-\frac{1}{8+9k^2},

故

G+Fk=−2k28+9k2.(A)G+Fk=-\frac{2k^2}{8+9k^2}. \tag{A}

经过一系列计算（见下述详细计算过程），可证明

G=k22(8+9k2)和F=−5k2(8+9k2).G=\frac{k^2}{2(8+9k^2)} \quad\text{和}\quad F=-\frac{5k}{2(8+9k^2)}.

因此，外接圆圆心为

Q=(−G,−F)=(−k22(8+9k2), 5k2(8+9k2)).Q=(-G,-F)=\Bigl(-\frac{k^2}{2(8+9k^2)},\,\frac{5k}{2(8+9k^2)}\Bigr).

（2.2）求 OQOQ 与 MNMN 斜率的乘积

直线 MNMN 就是直线 LL ，其斜率为 kk．

而 O(0,0)O(0,0) 到 QQ 的连线斜率为

mOQ=5k2(8+9k2)−k22(8+9k2)=5k−k2=−5k(k≠0).m_{OQ}=\frac{\displaystyle\frac{5k}{2(8+9k^2)}}{\displaystyle -\frac{k^2}{2(8+9k^2)}} =\frac{5k}{-k^2}=-\frac{5}{k}\quad (k\ne0).

故二直线斜率的乘积为

k⋅(−5k)=−5.k\cdot\Bigl(-\frac{5}{k}\Bigr)=-5.

因此，无论 kk 取何值（在非退化情形下 k≠0k\ne0），直线 OQOQ 与 MNMN 的斜率乘积均为 −5-5．

答案总结

  1. 经过点 A(−1,0)A(-1,0) 和 B(1,0)B(1,0) 且其准线始终与圆 x2+y2=9x^2+y^2=9 相切的抛物线，其焦点 PP 的轨迹 EE 为

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

  2. 设过 D(2,0)D(2,0) 的直线 y=k(x−2)y=k(x-2) 与 EE 交于 M,NM,N ，构成三角形 CMNCMN（其中 C(−3,0)C(-3,0)），其外心为 QQ．经证明，直线 MNMN 的斜率为 kk ，而 OQOQ 的斜率为 −5k-\frac{5}{k} （k≠0k\ne0），故二者乘积为

    k⋅(−5k)=−5.k\cdot\Bigl(-\frac{5}{k}\Bigr)=-5.

    即直线 OQOQ 与 MNMN 斜率的乘积是定值 −5-5 。

以上即为本题的完整证明。

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