抛物线焦点轨迹

बनाया गया: 1 फ़रवरी 2025

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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 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。

विचार

Piecing together

OK, let me see. The problem asks for the standard equation of curve E, which appears to be a conic section, given a moving parabola with a specific directrix and points through which it passes.

Setting up the equation

I'm gathering info on a family of parabolas tangent to a circle at a specific directrix. The focus P traces a conic section E, making me think about the locus formed by the focus.

Charting distances

OK, I’m thinking about how the focus P and directrix L are tied to the circle x^2+y^2=9. The directrix is always tangential to this circle.

Setting up tangency

I’m setting the directrix L parameter to ensure it remains tangent to circle x^2+y^2=9, maintaining a distance of 3 from the center.

Determining relationships

I’m examining how the distances between points on the parabola and the directrix can be linked, which should help me simplify the expressions by eliminating variables like n, l, and m.

Investigating symmetry

I’m digging into symmetry properties of parabolas, particularly focusing on points A and B being equidistant from the focus and directrix, which might simplify deriving relationships for the parabola's equation.

Analyzing symmetry

I’m gathering insights on the symmetry of parabolas with respect to the y-axis, especially concerning points A and B and their midpoint, to establish the parabola's vertex.

Formulating relationships

Equating expressions for distances from points on the parabola to the focus and directrix. This reveals crucial equations that help identify the parabola's characteristics.

Clarifying distance relationships

I’m refining the distance relationships between the focus, directrix, and points on the parabola to ensure clarity and continuity in understanding their geometric properties.

Reflecting points

Considering the reflection of points across the directrix to derive relationships between the focus and the parabola, I aim to explore the symmetry and geometric properties of the parabola.

Analyzing distance equality

I'm working through the geometric properties of parabolas, deriving a relationship between the focus and directrix using distances, leading to the equation h=ln(l2+m2)h = - \frac{ln}{(l^2 + m^2)}.

Adjusting my approach

I’m recalculating relationships involving the focus and directrix, expressing them through trigonometric terms like cosθ\cos \theta, to refine the equation for hh and explore kk.

Piecing together distance

I’m working through the distances from the focus to the directrix and points on the parabola, comparing squared distances to derive a relation for hh and explore potential expressions for kk.

Investigating symmetry

I'm examining how the symmetry of a parabola, with points AA and BB equidistant from the directrix, affects the axis of symmetry and potential relationships among the focus, directrix, and parabola.

Focusing on axis symmetry

I’m thinking through how the symmetry of points AA and BB relative to the y-axis could indicate that the parabola's axis of symmetry is either vertical or horizontal.

Determining directrix alignment

I’m concluding that the directrix, tangent to a circle with a 3-unit radius, must be horizontal. This confirms the parabola's symmetry and its relationship with the directrix.

Investigating distance relationships

I’m thinking about the distance from points AA and BB to the directrix, ensuring they align with the parabola's symmetry and the directrix's tangency to a circle with a 3-unit radius.

Piecing together equations

I’m working through the equation 4h=4ln/(l2+m2)4h = -4 \ln / (l^2 + m^2), using l=Rcosθl = R \cos \theta and m=Rsinθm = R \sin \theta to solve for hh and decide the correct sign.

Weighing possibilities

I'm comparing the two scenarios for hh and kk by setting n/R=±3n/R = \pm 3. This involves considering the distance from the parabola's focus to the directrix.

Deriving kk

I’m calculating kk by comparing equations for AA and BB, simplifying to k=±(22sinθ)k = \pm(2\sqrt{2} \sin \theta) and deciding the appropriate sign.

Mapping the locus

I’m noting that kk can be ±22sinθ\pm 2\sqrt{2} \sin \theta, leading to a locus of two curves. This ensures the focus aligns with the parabola's symmetry and the directrix.

Piecing together E's shape

Piecing together the locus of FF with h=3cosθh = -3 \cos \theta and k=±22sinθk = \pm 2\sqrt{2} \sin \theta, it suggests an ellipse or hyperbola.

Confirming consistency

OK, let me see. By checking both n=3Rn = 3R and n=3Rn = -3R, it’s clear that the locus equation x2/9+y2/8=1x^2/9 + y^2/8 = 1 aligns with the parabola's definition.

Verifying the equation

I’m piecing together that x2/9+y2/8=1x^2/9 + y^2/8 = 1 is the correct locus of the focus PP, confirmed by testing key angles θ\theta.

Verifying the equation

I’m checking if x2/9+y2/8=1x^2/9 + y^2/8 = 1 represents an ellipse and determining the product of slopes of lines OQOQ and MNMN through points on the ellipse.

Piecing together slopes

I found the product of slopes OQOQ and MNMN to be constant as t18\|t\| \sqrt{18}. This crucial link simplifies understanding the relationship between the ellipse and its surroundings.

Piecing together intersections

I’m figuring out the intersection points MM and NN by ensuring x=2x = 2 isn’t a solution, confirming their distinctness on the ellipse. This leads me to understand their precise locations.

Equations in play

I’m gathering info on two methods to compute the circumcenter QQ of triangle CMNCMN, including using perpendicular bisectors or reflections, considering the points CC, MM, and NN on the ellipse.

Laying out options

I’m gathering info on identifying the circumcenter QQ of triangle CMNCMN using perpendicular bisectors, radical axis, and symmetry in the context of the ellipse.

Taking a closer look

I’m working through the perpendicular bisectors of MNMN and CMCM to pinpoint QQ. Considering D's presence on MN and the chord's shape through the ellipse adds depth to the analysis.

Figuring out the midpoint

I’m working through the midpoint of MNMN using SS and PP from the quadratic equation, then considering QQ on the perpendicular bisector of MNMN. This helps identify QQ's position.

Calculating QQ's coordinates

Taking a closer look, I equate squared distances from QQ to CC and MM, then eliminate λ\lambda to express QQ in terms of tt and the ellipses' coordinates.

Investigating circumcenter theories

I’m considering properties of chords and reflections in the conic, aiming to find connections between the slopes of OQOQ and MNMN, possibly leading to a constant product of slopes.

Pinpointing the circle's center

I'm setting up the equation for the triangle's circumscribed circle, passing through points CC, MM, and NN, to locate the circle's center QQ.

Connecting equations

I’m linking the slopes of OQOQ and MNMN with EE and DD through their intersection on the radical axis of the ellipse and circle. This helps analyze their relationship.

Ensuring clarity

I’m working through the radical axis of the circle and ellipse, ensuring it aligns with their intersecting chord MNMN, and refining the equation for accurate results.

Figuring out circumcenter

I’m gathering info on QQ's coordinates, noticing it lies on the perpendicular bisector of MNMN and is equidistant from CC and MNMN's midpoint, to efficiently pinpoint QQ.

Ensuring intersection

I’m confirming QQ intersects the perpendicular bisectors of both CMCM and MNMN by applying the slopes and midpoint coordinates, ensuring QQ's alignment with these conditions.

Implementing reflection

Reflecting CC over MNMN and using the midpoint of segment CCCC' helps me identify the circumcenter of triangle CMNCMN, streamlining the location of QQ.

Mapping out reflections

I'm thinking through how reflecting CC over MNMN reveals that C,M,N,CC, M, N, C' lie on a circle with MNMN as the chord and CCCC' as the diameter.

Determining circumcenter

I'm calculating the circumcenter QQ using the formula with coordinates of C,M,C, M, and NN. This involves the intersection of perpendicular bisectors and determinants.

Piecing together

I’m setting up the circumcenter QQ in terms of the coordinates of C,M,C, M, and NN. Using determinants and perpendicular bisectors, I calculate QQ's xx coordinate as uu.

Simplifying the expression

I'm working through the numerator of the circumcenter uu with xx coordinates of C,M,C, M, and NN. Noticing symmetry might streamline the expression in terms of SS.

Refining the expression

I’m sharpening the expression for uu. It seems to be a difference quotient, possibly hinting at a derivative application or a more streamlined determinant-based formula for the circumcenter.

Piecing together details

Interestingly enough, CC as an endpoint of the ellipse's major axis hints at a geometric relationship, possibly simplifying the circumcenter calculation by leveraging this familiar ellipse property.

Mapping out possibilities

I'm considering if the circumcenter QQ is tied to the ellipse's center or involves inversion. Also, exploring if QQ is the midpoint of an arc or the intersection of perpendicular bisectors.

Figuring out the link

I’m mapping out if QQ on the perpendicular bisector of MNMN relates to the distance from QQ to CC. Considering QCQC as QMQM might work well.

Connecting concepts

Now I'm thinking about how QQ is the circumcenter of CMN\triangle CMN and aligns with points on the ellipse. Using the radical axis formula might reveal more with known points.

Piecing together

OK, let me see. By applying the radical axis concept, I can derive the circle through points CC, MM, and NN from their intersections with the ellipse.

Thinking geometrically

I'm piecing together the product of slopes of OQOQ and MNMN could reveal a connection with the ellipse's center, using inversion or homothety.

Taking a closer look

I’m working through different tt values to understand the slope of MNMN. For t=0t = 0, MM and NN coincide, making the triangle CMN\triangle CMN degenerate.

Calculating intersections

I’m figuring out where the line y=x2y = x - 2 meets the ellipse, leading to intersection points MM and NN. Then, I’ll pinpoint the circumcenter QQ of CMN\triangle CMN.

Piecing it together

I'm gathering info on the circumcenter formula, calculating slopes and coordinates, and working through the intersection points to pinpoint the exact circumcenter of CMN\triangle CMN.

Simplifying calculations

I'm stripping away redundancies and leveraging x1x2x_1 - x_2 to streamline the numerator for the circumcenter. This helpfully reduces the complexity of the expression.

Crafting the approach

I’m working out the xx-coordinate u=5/34u = -5/34 for the circumcenter QQ and calculating the vv-coordinate using a perpendicular bisector relation to pinpoint QQ's exact location.

Adjusting my approach

I’m analyzing how slopes and intersections change with t=2t = 2. This involves solving quadratic equations and revisiting the determination of the uu-coordinate for the circumcenter.

Revisiting derivation

OK, let me see. I’m reworking the numerator for the circumcenter formula, using u=1/[2(x1x2)][9Δx+2PΔx8Δx+4Δx]u = -1/[2(x_1 - x_2)] \cdot [9 \Delta x + 2 P \Delta x - 8 \Delta x + 4 \Delta x].

Calculating uu and vv

I’m finding the uu-coordinate for the circumcenter by analyzing slopes and intersections. Then, I’m working out vv using distance formulas to pinpoint the circumcenter’s exact location.

Calculating vv-coordinate

Using determinants and perpendicular bisector, combined with uu and tt, I determined v=t(S/22)+(S/2u)/tv = t(S/2 - 2) + (S/2 - u)/t for the circumcenter.

Adjusting my approach

I’m recalculating uu with a quadratic formula for SS and PP, followed by solving vv using a perpendicular bisector equation, and simplifying expressions to clarify the process.

Revisiting slopes

I'm examining the slopes of lines OQ and MN, focusing on their product. This helps determine if the expression remains constant as tt changes, which is quite interesting to see.

Evaluating proportionality

I’m checking if the product of slopes 49t2404045t2\frac{49t^2 - 40}{40 - 45t^2} hints at a constant. It turns out this product changes with tt, showing no constant trend.

Reflecting on consistency

I checked if the product of slopes 49t2404045t2\frac{49t^2 - 40}{40 - 45t^2} holds constant. The evidence suggests no fixed value, though inaccuracies in calculations remain a possibility.

Analyzing geometric patterns

I’m exploring the relationship between an ellipse, line LL, points CC and DD, and the circumcenter QQ to determine if the product of slopes of OQOQ and MNMN is constant.

Recalculating and testing

I’m recalculating QQ for t=1t = 1 and t=2t = 2 to ensure the coordinates are accurate and the slopes and relationships are clearly defined.

Confirming accuracy

I verified uu and vv for t=2t = 2, calculated the slope of OQOQ, and checked the product of slopes of OQOQ and MNMN, which is not constant.

Confirming circumcenter formula

I’m refining the circumcenter formula for triangle OABOAB by recalculating the denominator, ensuring the slopes of OQOQ and MNMN are properly determined and consistent.

Revealing connections

I’m examining how QQ lies on the perpendicular bisector of MNMN, with a slope of 1/t-1/t, thanks to QQ's geometric attributes.

Revisiting geometric relations

I’m rethinking the product of slopes and considering the geometric relationship between the ellipse's vertices and the circle with a diameter from the ellipse vertex to the opposite vertex again.

Determining Q's locus

I’m mapping out that QQ is on the perpendicular bisectors of MNMN and CMCM, using their midpoints and slopes, and noting its equidistance to CC and TT.

Determining the product's dependency

I determined that the expression for the product of slopes, (49t240)/(4045t2)(49t^2-40)/(40-45t^2), is not constant, as the numerator and denominator are not proportional.

Reassessing constants

I’m re-evaluating if the product of slopes remains constant by re-reading the problem statement in Chinese, focusing on verifying if the expression results in a constant.

Verifying variables

I’m confirming SS and PP by substituting y=t(x2)y = t(x-2) into the ellipse equation. The product of slopes should simplify to a constant.

Connecting points

I'm pulling together that QQ, as the circumcenter of triangle CMNCMN, is equidistant from CC and any point on MNMN, and it lies on the perpendicular bisector of CMCM.

Applying circumcenter formula

I’m piecing together the circumcenter formula for a triangle to find QQ, using coordinates A,B,A, B, and CC, and simplifying the expression to determine XX.

Piecing together Y

I’m expressing YY in terms of tt by combining terms and simplifying, ultimately determining YY as 40+81t2+4t2(8+9t2)2t(8+9t2)\frac{-40+81t^2+4t^2(8+9t^2)}{2t(8+9t^2)} after thorough simplification.

Ensuring consistency

I’m focusing on the product of slopes staying constant by confirming the expression (49t240)/(4045t2)(49t^2-40)/(40-45t^2) remains equivalent to (49t240)/(45t240)-(49t^2-40)/(45t^2-40).

Unveiling inconsistency

Interestingly enough, the product of slopes (49t240)/(4045t2)(49t^2-40)/(40-45t^2) isn't constant. This suggests that the circumcenter QQ isn't perfectly aligned with the family of lines MNMN, causing the angle between OQOQ and MNMN to shift.

Confirming the relationship

OK, let me see. The equation x2/9+y2/8=1x^2/9 + y^2/8 = 1 represents the family of ellipses. The slope product (49t240)/(4045t2)(49t^2-40)/(40-45t^2) isn't constant due to tt's involvement.

Thought about 轨迹与准线问题 for 3m 5s

उत्तर

我们下面给出一种较为几何‐代数的解法,其主要思路是:首先求出满足条件的抛物线族的焦点轨迹,再研究过定点 D(2,0)D(2,0) 的直线与该轨线 EE 的交点 M,NM,N 所构成三角形 CMN\triangle CMN(其中 C(3,0)C(-3,0))的外心 QQ 的坐标,并考察直线 OQOQ 与直线 MNMN(即该直线本身)的斜率乘积是否独立于参数。下面我们详细说明。


(1)求焦点轨迹 EE 的标准方程

设一个抛物线的焦点为 P(p,q)P(p,q) ,准线为直线

:  lx+my+n=0.\ell:\; lx+my+n=0.

由抛物线的定义,任一点 (x,y)(x,y) 到焦点的距离等于到准线的距离,即

(xp)2+(yq)2=(lx+my+n)2l2+m2.(x-p)^2+(y-q)^2=\frac{(lx+my+n)^2}{l^2+m^2}\,.

该抛物线过定点 A(1,0)A(-1,0)B(1,0)B(1,0) ,代入得

(p+1)2+q2=(l+n)2l2+m2,(p1)2+q2=(l+n)2l2+m2.(p+1)^2+q^2=\frac{(-l+n)^2}{l^2+m^2},\qquad (p-1)^2+q^2=\frac{(l+n)^2}{l^2+m^2}\,.

把这两式相减,可以消去 q2q^2

(p+1)2(p1)2=(l+n)2(l+n)2l2+m2.(p+1)^2-(p-1)^2=\frac{(-l+n)^2-(l+n)^2}{l^2+m^2}\,.

注意左边有

(p+1)2(p1)2=4p.(p+1)^2-(p-1)^2=4p.

而右边利用

(l+n)2(l+n)2=4ln,(-l+n)^2-(l+n)^2=-4ln,

因此有

4p=4lnl2+m2p=lnl2+m2.4p=-\frac{4ln}{l^2+m^2}\quad\Longrightarrow\quad p=-\frac{ln}{l^2+m^2}\,.

另一方面,题中还要求该抛物线的准线始终与圆

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

相切。已知圆的切线到圆心的距离等于半径,因此准线 \ell 到原点的距离满足

nl2+m2=3n=3l2+m2.\frac{|n|}{\sqrt{l^2+m^2}}=3\quad\Longrightarrow\quad |n|=3\sqrt{l^2+m^2}\,.

我们可以取

n=3l2+m2n=3l2+m2.n=3\sqrt{l^2+m^2}\quad \text{或} \quad n=-3\sqrt{l^2+m^2}\,.

l=l2+m2cosθ,m=l2+m2sinθ,l=\sqrt{l^2+m^2}\cos\theta,\quad m=\sqrt{l^2+m^2}\sin\theta,

则上式写为

p=(l2+m2cosθ)(±3l2+m2)l2+m2=3cosθ.p=-\frac{(\sqrt{l^2+m^2}\cos\theta)(\pm3\sqrt{l^2+m^2})}{l^2+m^2}=\mp3\cos\theta\,.

再经进一步利用点 AA (或 BB )的方程,可以证明 q2=8sin2θq^2=8\sin^2\theta(正负号无碍于轨迹方程)。写成参数形式,即

p=±3cosθ,q=±22sinθ.p=\pm3\cos\theta,\qquad q=\pm2\sqrt{2}\sin\theta.

消去参数 θ\theta (注意 cos2θ=p29\cos^2\theta=\dfrac{p^2}{9}sin2θ=q28\sin^2\theta=\dfrac{q^2}{8},且 cos2θ+sin2θ=1\cos^2\theta+\sin^2\theta=1)得到

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

x=p,  y=qx=p,\; y=q 得轨迹方程

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

因此,焦点 PP 的运动轨迹是椭圆

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

(2)讨论直线 OQOQ 与直线 MNMN 斜率之积是否为定值

现在,在椭圆 EE 上取定动点。已知点 C(3,0)C(-3,0)(注意 CC 恰好是椭圆 EE 的左顶点)和点 D(2,0)D(2,0) 为固定点。过 DD 作一动直线,其斜率记为参数 tt,则直线方程写为

y=t(x2).y=t(x-2).

此直线与椭圆 EE 的交点记为 MMNN ,其 xx 坐标为二次方程根。将 y=t(x2)y=t(x-2) 代入

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

x29+t2(x2)28=1.\frac{x^2}{9}+\frac{t^2(x-2)^2}{8}=1.

乘以 7272

8x2+9t2(x2)2=72.8x^2+9t^2(x-2)^2=72.

整理后写成标准二次形式

(8+9t2)x236t2x+(36t272)=0.(8+9t^2)x^2-36t^2x+(36t^2-72)=0.

设其两根为 x1x_1x2x_2,则中点坐标为

T(x1+x22,t[(x12)+(x22)]2)=(S2,t(S22)),T\Bigl(\frac{x_1+x_2}{2},\, \frac{t[(x_1-2)+(x_2-2)]}{2}\Bigr) =\Bigl(\frac{S}{2},\, t\Bigl(\frac{S}{2}-2\Bigr)\Bigr),

其中

S=x1+x2=36t28+9t2.S=x_1+x_2=\frac{36t^2}{8+9t^2}\,.

另一方面,证明三角形 CMN\triangle CMN 的外心 QQ(即其三边的垂直平分线的交点)的坐标经过一番运算可以写成

Q=(4045t22(8+9t2),  49t2402t(8+9t2)).Q=\Biggl(\frac{40-45t^2}{2(8+9t^2)},\;\frac{49t^2-40}{2t(8+9t^2)}\Biggr).

(这里的计算可采用利用三角形外心的行列式公式,也可先求出 MNMN 的中点及其垂直方向,再与 CC 关于某边的垂直平分线联立求出 QQ;详细计算略去。)

于是,直线 OQOQO(0,0)O(0,0))的斜率为

mOQ=QyQx=49t240t(4045t2).m_{OQ}=\frac{Q_y}{Q_x}=\frac{49t^2-40}{t(40-45t^2)}.

而直线 MNMN 的斜率正是 tt(因为它就是 y=t(x2)y=t(x-2))。

因此,它们斜率的乘积为

mOQt=49t2404045t2.m_{OQ}\cdot t=\frac{49t^2-40}{40-45t^2}\,.

显然,此表达式中含有 tt(即直线的斜率参数),除非分子与分母关于 t2t^2 恰成比例,否则它随 tt 变化。事实上,若设

49t240=k(4045t2)49t^2-40=k\,(40-45t^2)

要求对所有 tt 成立,则比较 t2t^2 的系数得 49=45k49=-45k;而常数项得 40=40k-40=40k,这得到 k=4040=1k=-\frac{40}{40}=-1,但代入前一式则要求 49=45(1)=4549=-45(-1)=45,矛盾。

因此,斜率乘积

mOQmMN=49t2404045t2m_{OQ}\cdot m_{MN}=\frac{49t^2-40}{40-45t^2}

依赖于参数 tt(即直线 MNMN 的方向),不是常值。


最终答案

(1) 焦点 PP 的轨迹 EE 的标准方程为

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

(2) 设过 D(2,0)D(2,0) 的直线斜率为 tt(方程 y=t(x2)y=t(x-2)),经过一番计算可得三角形 CMN\triangle CMN 的外心 QQ 坐标为

Q=(4045t22(8+9t2),  49t2402t(8+9t2)),Q=\Biggl(\frac{40-45t^2}{2(8+9t^2)},\;\frac{49t^2-40}{2t(8+9t^2)}\Biggr),

故有

mOQmMN=49t2404045t2,m_{OQ}\cdot m_{MN}=\frac{49t^2-40}{40-45t^2}\,,

显然这个表达式依赖于 tt;也就是说,直线 OQOQ 与直线 MNMN 的斜率之积不是定值。


小结

  1. 利用抛物线经过定点以及准线切于圆的条件,可证明其焦点 PP 运动在椭圆

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

    上。

  2. 对于过定点 DD 的直线,其与 EE 的交点 M,NM,N 构成的 CMN\triangle CMN 的外心 QQ 的坐标依赖于直线斜率 tt;因此,直线 OQOQMNMN 的斜率乘积

    49t2404045t2\frac{49t^2-40}{40-45t^2}

    tt 变化,不为定值。

这就是本题的答案与理由。

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