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 + ...

प्रश्न

已知过点 $A(-1, 0)$ 、 $B(1, 0)$ 两点的动抛物线的准线始终与圆 $x^2 + y^2 = 9$ 相切，该抛物线焦点 $P$ 的轨迹是某圆锥曲线 $E$ 的一部分。<br>(1) 求曲线 $E$ 的标准方程；<br>(2) 已知点 $C(-3, 0)$ ， $D(2, 0)$ ，过点 $D$ 的动直线与曲线 $E$ 相交于 $M$ 、 $N$ ，设 $\triangle CMN$ 的外心为 $Q$ ， $O$ 为坐标原点，问：直线 $OQ$ 与直线 $MN$ 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

Solving parabola problem

I need to solve this in Chinese, focusing on a parabola with fixed focal chord AB, where A(-1,0) and B(1,0), and the directrix is tangent to the circle x^2 + y^2 = 9. The focus P moves, and I'm asked to find the equation of curve E formed by the set of all such foci. The directrix is tangent to the circle, and I need to find the relationship between the coordinates of the focus. The focus's position involves using the definition of a parabola: the distance to the focus equals the distance to the directrix.

Considering distances and focus

We have two fixed points A(-1,0) and B(1,0) on the parabola. For a point P (the focus) and a directrix L, the distance from A to P equals the distance from A to L, and similarly for B. I’m thinking of using the equidistant property of parabolas and the relationship between focus and directrix to determine how A and B connect to the focus. The directrix L is tangent to the circle x^2 + y^2 = 9, and we know that this distance is 3. This can help solve the problem.

Deriving focus and directrix relations

I’m working through relationships involving the focus and directrix of the parabola. For A(-1,0) and B(1,0), the distance from A and B to the directrix involves calculating absolute values based on the equation of the line. When we subtract these distances, I find a relation between the coefficients a and the focus p. After some more steps, it seems the focus lies on the y-axis due to the symmetry of A and B. Contradictions suggest corrections, and I suspect the focus is at (0, q), but the actual solution depends on testing specific cases.

Locus of the focus

I'm finding the locus of the focus for the parabola with directrices at y = 3 or y = -3, and passing through A(-1, 0) and B(1, 0). After some calculations, I derive that the locus of foci satisfies the ellipse equation $\frac{x^2}{9} + \frac{y^2}{8} = 1$ . The focus points lie on this ellipse, with $p^2 + q^2$ between 8 and 9. So, the answer is an ellipse, and I confirm the equation matches the necessary conditions.

Finding circumcenter Q

I'm exploring how to calculate the circumcenter Q for triangle CMN, where points M and N are on the intersection of the ellipse and a line. I want to find the relationship between the coordinates of Q, the midpoint of M and N, and the fixed point C. To do this, I use the fact that Q is equidistant from C and any M or N. I’m trying to eliminate unnecessary variables, and I'm also considering using certain geometrical properties like the radical axis to simplify the process.

None

Reasoned for 1m 17s

उत्तर

下面给出一种答案思路，最终结果为

焦点的轨迹 $E$ 的标准方程为

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

对于点 $C(-3,0)$ 和 $D(2,0)$ 以及任一过 $D$ 的直线（设斜率为 $k$ ）与 $E$ 的交点 $M,N$ 构成的三角形 $CMN$ ，其外心 $Q$ 与原点 $O$ 所连直线的斜率与直线 $MN$ 的斜率的乘积恒等于

-5.

下面简单介绍一下证明思路。

（1）求 $E$ 的方程

设动抛物线的焦点为 $P=(p,q)$ ；该抛物线的准线为直线 $L$ （一般式可写为

ax+by+c=0,\quad a^2+b^2=1,

且因 $L$ 与圆 $x^2+y^2=9$ 相切，有 $|c|=3$ ）。

因为抛物线定义为：任一点到焦点的距离等于到准线的距离，所以两固定点 $A(-1,0)$ 和 $B(1,0)$ 都满足

\sqrt{(p\pm 1)^2+q^2}=\bigl|\,\mp a+c\,\bigr|.

先将这两式相减，便可得

4p=-4ac\quad\Longrightarrow\quad ac=-p.

再将两式相加，经化简可得

a^2=p^2+q^2-8.

利用 $a^2+b^2=1$ 以及 $c^2=9$ ，可以推出

p^2+q^2\le 9\quad\text{且}\quad p^2+q^2\ge 8.

消去参数 $a$ 后，经过计算可得

8p^2+9q^2=72,

即

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

记 $x=p,\; y=q$ 即得所求圆锥曲线 $E$ 的标准方程。

（2）求直线 $OQ$ 与 $MN$ 斜率乘积

取过 $D(2,0)$ 的直线 $L:\; y=k(x-2)$ （ $k$ 为参数），该直线与 $E$ 相交于 $M,N$ 。证明中经过较多代数运算，可以证明：若记 $Q$ 为三角形 $CMN$ 的外心，则用坐标计算可证明

m(OQ)\cdot m(MN)=\frac{\mu}{\lambda}\,k=-5,

其中 $(-\lambda/2,\,-\mu/2)$ 为该圆（ $CMN$ 的外接圆）的圆心。也就是说，不论直线 $L$ 如何选取（只要过 $D$ ），都有

m(OQ)\,m(MN)=-5.

综上所述，答案为

焦点轨迹为

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

对任意取过 $D(2,0)$ 的直线，其与 $E$ 的交点 $M,N$ 所构成的三角形 $CMN$ 的外心 $Q$ 与原点连线与直线 $MN$ 的斜率乘积恒为 $-5$ .

प्रश्न

विचार

उत्तर

（1）求 EEE 的方程

（2）求直线 OQOQOQ 与 MNMNMN 斜率乘积

इस प्रश्नोत्तर को साझा करें

（1）求 $E$ 的方程

（2）求直线 $OQ$ 与 $MN$ 斜率乘积