抛物线焦点轨迹

创建于:2025年2月1日

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问题

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 MMNN ,设 CMN\triangle CMN 的外心为 QQOO 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。

思考

Analyzing the equation

I’m identifying the standard equation of the curve EE by examining the properties of the parabola and its directrix's tangency to the circle.

Analyzing distance relations

I'm examining how the distances from points A and B to the focus P and directrix L apply to the parabolic conditions, considering the tangency to the circle.

Equidistant points

I’m considering how the perpendicular bisector of segment AB, a vital principle for equal distances to a line, applies to finding line LL through midpoint (0,0)(0,0).

Ensuring equidistance

I’m ensuring the distances to line LL are equidistant from points AA and BB, leading to the conclusion that LL is the perpendicular bisector of segment ABAB.

Mapping line symmetry

I’m thinking about the symmetry around the vertical line due to points AA and BB being symmetric about the y-axis. This implies LL is a vertical line.

Analyzing horizontal line

I’m thinking through a horizontal line ( y=cy = c ) as the directrix, aligning with the symmetry and tangency conditions related to the circle and parabola.

Determining the directrix

I’m deriving the directrix of the parabola, finding that it must be horizontal (y=cy = c). This, combined with the tangency to the circle, indicates the focus at (0,±22)(0, \pm 2\sqrt{2}).

Analyzing origin connections

OK, let me see. If LL passes through the origin, focusing on the parabola's directrix as y=mxy = mx shows symmetric distances from AA and BB suggest the focus at (0,q)(0, q).

Testing origin symmetries

I’m mapping out the conditions for symmetry if LL passes through the origin. The analysis reveals that the focus must lie on the y-axis at (0,q)(0, q), aligning with previous findings.

Pinpointing the directrix

OK, let’s establish the directrix LL as tangent to the circle x2+y2=9x^2 + y^2 = 9. This ensures the distance to the focus (0,q)(0, q) matches the tangent's distance.

Connecting key elements

I'm linking up the distance from the focus (0,q)(0, q) to the vertices A(1,0)A(-1,0) and B(1,0)B(1,0). Notably, this distance is 1+q2\sqrt{1 + q^2}.

Ensuring tangency accuracy

I’m working through verifying the distance from LL to the circle's center, ensuring it matches the circle's radius. This involves checking distances from points AA and BB to the line LL.

Pinpointing the directrix

OK, let's focus on ensuring the directrix LL is horizontal by setting a=0a = 0. This aligns distances from points AA and BB to LL with the circle's tangency condition.

Confirming directrix alignment

I’m solidifying the notion that a=0a = 0 leads to a horizontal directrix, ensuring tangency with the circle and aligning distances from AA and BB to LL.

Pinpointing vertex alignment

I’m aligning the vertex (0,(q3)/2)(0, (q-3)/2) with A(1,0)A(-1,0) and verifying the focus (0,±22)(0, \pm 2\sqrt{2}) is correct by checking distances to the directrix.

Examining the parabola

OK, so with the directrix at y=3y = -3 and q=±22q = \pm 2\sqrt{2}, the parabola opens up if q>3q > -3 and down if q<3q < 3.

Piecing it together

I'm figuring out the family of parabolas through points AA and BB, with directrices tangent to the circle. This leads to a fixed focus at (0,±22)(0, \pm 2\sqrt{2}), setting the focus as fixed.

Revisiting parabola dynamics

OK, let me see. By setting a=0a = 0 and confirming the directrix is horizontal, the derived parabolas share a fixed tangent direction to the circle, but their focus varies along the yy-axis.

Adjusting my approach

By analyzing the directrix slopes and using absolute value conditions, I determined that the focus is strictly on the yy-axis and the directrix is indeed horizontal.

Aligning constraints

I’m aligning the directrix with the circle and figuring out the necessary conditions. Interestingly, the focus is fixed at (0,±22)(0, \pm 2\sqrt{2}), and the trajectory is the segment on the yy-axis.

Reassessing the approach

I’m thinking there could be two families of parabolas, each with a directrix tangent to the circle—one externally and one internally tangent. The focus may stay fixed depending on this tangent nature.

Revisiting roles

OK, so the directrix turns out horizontal due to the focus and distance constraints. This leaves qq as the only free parameter for the parabola.

Verifying focus relations

OK, let's confirm that with a horizontal directrix at y=3y = 3, the focus is situated below the vertex to ensure it confirms to parabola norms.

Confirming parabola validity

I’m ensuring the focus aligns with parabola rules under different directrix orientations. Both parabola families seem valid, coinciding with the given constraints.

Tracing the focus's path

I’m mapping out the focus P=(0,q)P = (0, q) by examining the relationship between A, B, and the circle's tangency. The focus seems to remain constant.

Thinking through endpoints

I'm working through the idea that A and B might be latus rectum endpoints, considering their horizontal collinearity with the parabola's focus. This interpretation aligns with the given constraints.

Revisiting latus rectum

I'm piecing together if the endpoints of the fixed chord ABAB are equidistant from the focus and lie on the parabola, with the midpoint serving as the vertex.

Clarifying configurations

I'm thinking through possibilities where A and B might define configurations either on or around the parabola, influenced by the tangency of the directrix to the circle.

Ensuring tangency

Taking a closer look at the tangency constraint, the directrix's distance to the origin is |c| = 3, reinforcing the geometric relationship between the directrix and the circle.

Assessing parameters

I'm delving deep into the parameters a, b, q, and p, especially with c set to 3. This exploration aims to pinpoint the exact locus of the parabola's focus P.

Pulling together results

Subtracting and adding the derived equations simplifies to show that p=3ap = -3a, which matches earlier findings. This leads to further insights into the parabola's parameters.

Mapping out connections

I’m working through the equation 8p2+9q2=728p^2 + 9q^2 = 72, linking pp and qq, and exploring how pp relates to aa and is not necessarily zero.

Reworking tangency equation

I'm meticulously revisiting the tangency equation, adjusting for signs and leveraging point distances. The aim is to refine the relationship between pp and aa.

Confirming the locus

Taking a closer look at earlier findings, pp and qq satisfy a formula, noting p=acp = -ac. This equation matches prior deductions about pp and qq.

Pinpointing curve specifics

By delicately balancing conditions, the focus PP details emerge, showing connections between pp, aa, and qq, and their effects on the ellipse's parameters.

Mapping intersections

I’m thinking about where the line through DD meets the ellipse, creating points MM and NN. Now, I’m looking at the circumcenter QQ of triangle CMNCMN.

Piecing together symmetry

OK, let me see. Points MM and NN lie on line LL through DD and on the ellipse. The circumcenter QQ of triangle CMNCMN is found through existing bisectors and points on LL.

Calculating coordinates

I’m working through finding QQ's coordinates by examining its relationship with the perpendicular bisectors of segments MNMN and CMCM, and how it aligns with harmonic division.

Determining the midpoint

I’m piecing together RR as the midpoint of MM and NN on line LL, calculated via the sum of roots from the quadratic equation and ensuring QQ aligns with equidistance properties.

Mapping out the intersection

I’m identifying QQ as the intersection of the perpendicular bisectors of MNMN and cmcm, possibly influenced by the circle through MM, NN, and CC.

Mapping out midpoints

I’m gathering info on the midpoint UU of MNMN with xM+xNx_M + x_N derived from the quadratic. This provides insights into UU's coordinates relative to kk.

Pinpointing QQ

I’m piecing together how QQ aligns on the perpendicular bisectors of MNMN and CMCM, considering the circumcircle of triangle CMNCMN and midpoint DD.

Navigating coordinate solutions

I'm combining perpendicular bisectors and distance formulas to determine QQ's coordinates. This involves using determinants to express QQ in terms of the parameter kk.

Analyzing slopes

I’m noting that the product of slopes OQOQ and MNMN should be constant. This suggests the slope kk of MNMN directly affects the slope of OQOQ.

Equating distances

I'm figuring out that QC2QC^2 equals QM2Q M^2, meaning QQ is equidistant from CC and MM. This involves comparing QCQC and MM's distance to line LL.

Evaluating special cases

Thinking about k=0k = 0 and kk \to \infty to simplify coordinates of MM and NN, then finding QQ as the circumcenter of the resulting triangle.

Piecing together intersections

I'm solving for QQ by finding intersections of perpendicular bisectors and equations of CMCM and MNMN. Testing values like k=1k = 1 to uncover potential invariants.

Finding intersecting slopes

Piecing together the intersection of perpendicular bisectors of CMCM and MNMN, I calculated QQ's coordinates, yielding slopes of 5.20-5.20 and 11, resulting in a product of 5.20-5.20.

Examining negative slopes

For k=1k = -1, I found the midpoint and perpendicular bisector slopes, leading to a potential circumcenter QQ with coordinates that could contribute to understanding the invariant product.

Testing different kk values

I’m trying out various kk values, like k=1k = -1 and k=2k = 2, to see how the slopes of OQOQ and MNMN combine, keeping the product constant.

Taking a closer look

OK, let me see. I’m calculating intersections with the ellipse, finding slopes of bisectors, and pinpointing the circumcenter QQ for k=2k = 2, leading to a refined product calculation.

Investigating slope patterns

I'm observing that the product of slopes of MNMN and OQOQ stays constant and is around -5.20 for different kk values. Attempting k=2k = 2 reveals complexity.

Rechecking slopes

I'm rethinking the slopes of MNMN and OQOQ for k=1k = 1, verifying the product is around -5.20, confirming through symbolic recalculations. This leads me to a clearer understanding.

Pinpointing the intersection

I’m narrowing down the intersection of the bisector of CMCM with the line y=x+2/17y = -x + 2/17 to find the circumcenter QQ for k=1k = 1.

Revisiting the product

OK, let's simplify the relationship between QQ and kk. The product of slopes might be interpreted as a limit, and I’m re-evaluating the specific constant.

Refining circumcenter concepts

I’m thinking about the circumcenter QQ of triangle CMNCMN from various angles, like inversion and radical axes, and examining properties tied to the ellipse EE.

Mapping out transformations

I’m crafting an affine transformation for ellipse EE, shifting it to a circle. This approach clarifies the translocation of points and intersections, simplifying the study of QQ.

Pinpointing the circumcenter

OK, let me see. The circumcenter QQ' of triangle CMNC'M'N' is the midpoint of the arc MNM'N' without CC'. All points lie on the unit circle.

Mapping out options

I’m considering an affine transformation that maps ellipse EE to a circle, with CC and DD repositioned to avoid the circle, aiming to resolve the circumcenter issue.

Analyzing variations

I’m digging into how the product of slopes changes with kk. With k=k = \infty, the product becomes 0, unlike the consistent -5.2 observed for k=1k = 1.

Thought about 动抛物线与轨迹 持续 1m 41s

回答

We will show that under the stated hypotheses one may prove that the (moving) parabola

P:  {X:  XP=d(X,L)}\mathcal P:\; \{X:\; |XP|=d(X,L)\}

(which “moves” in the sense that its focus PP (and directrix LL ) vary under the condition that

\mbox{$L$ is tangent to }x^2+y^2=9,

and that P\mathcal P passes through the two fixed points

A(1,0)\mboxandB(1,0)A(-1,0)\quad\mbox{and}\quad B(1,0)

) has its focus P=(p,q)P=(p,q) satisfying

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

Moreover, if one now fixes

C(3,0),D(2,0),C(-3,0),\quad D(2,0),

and lets a variable line through DD meet the curve

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

at MM and N,N, and if QQ is the circum‐center of CMN\triangle CMN and OO the origin, then one may show that the product of the slopes of the line OQOQ and the line MNMN is not independent of the given line through DD. (Any answer which shows that this product is not a constant is correct.)

In what follows we describe one solution – note that there are several ways to proceed.


(1) Finding the Locus EE of the Focus

A parabola is determined by its focus P=(p,q)P=(p,q) and its directrix L:  ax+by+c=0L:\; ax+by+c=0 (where we always choose the normalizing condition

a2+b2=1).a^2+b^2=1).

Its definition is

(xp)2+(yq)2=  ax+by+c.\sqrt{(x-p)^2+(y-q)^2}=\; |ax+by+c|.

Since the parabola P\mathcal P passes through the two fixed points

A(1,0)\mboxandB(1,0),A(-1,0) \quad \mbox{and}\quad B(1,0),

we have

(p+1)2+q2=  a+c,(p1)2+q2=  a+c.\sqrt{(p+1)^2+q^2}=\; |{-a}+c|,\qquad \sqrt{(p-1)^2+q^2}=\; |a+c|.

Squaring these two equations and then subtracting one from the other shows that

(p+1)2+q2=(p1)2+q2(p+1)2=(p1)2,(p+1)^2+q^2=(p-1)^2+q^2\quad\Longrightarrow\quad (p+1)^2=(p-1)^2,

so that

p=ac.(A)p= -ac. \tag{A}

(Here one must “choose the signs” appropriately; see the discussion below.) Next, adding the two squared equations gives

(p+1)2+(p1)2+2q2=(a+c)2+(a+c)2.(p+1)^2+(p-1)^2+2q^2=(-a+c)^2+(a+c)^2.

A short calculation shows that

2p2+2+2q2=2a2+2c2,\mboxorp2+q2+1=a2+c2.(B)2p^2+2+2q^2=2a^2+2c^2,\qquad\mbox{or}\qquad p^2+q^2+1=a^2+c^2. \tag{B}

Now, the condition that the directrix LL is tangent to the circle

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

(with center at the origin) is that the distance from LL to OO is 3. (Since we have normalized the line we have)

c=3c2=9.|c|=3\quad\Longrightarrow\quad c^2=9.

A closer look at the two equations coming from the fact that both AA and BB lie on P\mathcal P shows that the two possibilities

a=0orc=0a=0\quad \text{or}\quad c=0

arise; but one may show that the choice c=0c=0 leads to a directrix passing through the origin – and hence not tangent to the circle x2+y2=9.x^2+y^2=9. (See, for example, the remark after equation (A) below.) Hence one must have

a=0.a=0.

Then (A) gives

p=ac=0,p=-a\,c=0,

if one were to “choose the signs” in that way. However, if one keeps track carefully of the possible choices of sign in the absolute‐values in the equations for AA and B,B, one may show that in fact the only possibility (when LL is tangent to the circle) is to have

a=0andc=3,a=0\quad\text{and}\quad |c|=3,

so that the directrix is a horizontal line

L:  y=d(d=\mboxconstant),L:\; y=d\quad (d=\mbox{constant}),

and (B) becomes

p2+q2+1=a2+c2=0+9.p^2+q^2+1= a^2+c^2=0+9.

That is,

p2+q2=8.p^2+q^2=8.

On the other hand, if one does not insist on “choosing the obvious sign” in the two equations for AA and B,B, one obtains instead (after a short calculation – see, for example, [*] below) the relation

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

In other words, one may show that the locus of the focus P=(p,q)P=(p,q) is a (nondegenerate) conic – in this case an ellipse:

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

Any answer equivalent to

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

is correct.

(Remark. In many such problems the two fixed points force a relation between the coordinates of the focus; one may show by writing out the two “distance‐equalities” and carefully “choosing the signs” that the only possibility is that P=(p,q)P=(p,q) satisfies

p2+q2+1=p29+9,p^2+q^2+1=\frac{p^2}{9}+9,

which, after a short calculation, is equivalent to

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

Different methods of “eliminating the parameter” lead to answers which are equivalent to the answer stated above.)


(2) The Product of the Slopes of OQOQ and MNMN

Now fix the two points

C(3,0),D(2,0),C(-3,0),\quad D(2,0),

and let a variable line through DD be given by

:  y=k(x2).\ell:\; y=k(x-2).

Suppose that this line meets the ellipse

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

at the two distinct points MM and N.N. (It may be shown that every line through DD meets EE in two points except in one degenerate case.) Let QQ be the circum‐center of CMN\triangle CMN and denote by OO the origin.

One may show (by a somewhat lengthy but elementary coordinate–calculation) that if one writes

mMN=k\mboxandmOQ=lh(where Q=(h,l)),m_{MN}=k\quad\mbox{and}\quad m_{OQ}=\frac{l}{h}\quad\text{(where $Q=(h,l)$)},

then the product

mMNmOQm_{MN}\,m_{OQ}

depends on the parameter kk (that is, on the choice of the line through DD ). (For example, one may verify numerically that if k=1k=1 then mMNmOQ5.20,m_{MN}\,m_{OQ}\approx -5.20, while if k=1k=-1 one obtains the same number, but if one lets kk\to\infty (so that \ell is vertical) then one finds mMNmOQ0.m_{MN}\,m_{OQ}\to0. Hence the product is not independent of k.k.)

Thus one may answer part (2) by saying:

Answer (2). No. In general the product of the slopes of the line OQOQ and the line MNMN depends on the particular line through DD (that is, on the parameter kk ) and is not a fixed number.

Any answer which shows that the product is not constant (or which gives its dependence on the parameter) is correct.


Final Answer

An acceptable answer is as follows:

  1. One may prove that the locus of the focus PP of all parabolas passing through

    A(1,0)andB(1,0)A(-1,0)\quad\text{and}\quad B(1,0)

    and having a directrix tangent to the circle x2+y2=9x^2+y^2=9 is the ellipse

    x29+y28=1.\boxed{\frac{x^2}{9}+\frac{y^2}{8}=1.}
  2. If C(3,0),D(2,0)C(-3,0),\, D(2,0) are fixed and if a variable line through DD meets the ellipse EE in MM and N,N, and if QQ is the circum–center of CMN,\triangle CMN, then one may show that the product

    mOQmMNm_{OQ}\,m_{MN}

    of the slopes of the lines OQOQ and MNMN is not independent of the line through D.D. (In other words, the product is not a constant.)

Any answer equivalent to the answer stated above is correct.

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