Is there a point P on the parabola's axis of symmetry such that ∠APB = 2∠AQB and there is only one s

616947616 shared this question 3 years ago

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The parabola y = a (x + 2) ^ 2 + c intersects the X axis at A and B, and intersects the Y axis at C. We know that point A (-1,0), OB = OC.

(1) Write the analytical expression of the parabola

(2) Point Q is a point on the straight line y = -x-4. Is there a point P on the parabola's axis of symmetry such that ∠APB = 2∠AQB and there is only one such point Q? If present, the coordinates of the point P are requested. If not, please explain why.

Files: 11.png

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mathmagic ● 3 years ago

distance((x0,y0),(x1,y1))=sqrt((x1-x0)^2+(y1-y0)^2) then repeat:

write and solve the bold equation in my post

distance(P,B)=distance(P,Q) P=(-2,y) Q=(-t-4,t) then y=(2t^2+4t+3)/(2t)

https://help.geogebra.org/t...

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616947616 ● 3 years ago

Correct the title

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(2) Point Q is a point on the straight line y = -x-4. Is there a point P on the parabola's axis of symmetry such that ∠APB = 2∠AQB and there is only one such point Q? If it exists, find the coordinates of point P. If not, please explain why.

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mathmagic ● 3 years ago

 Is there a point P on the parabola's axis of symmetry such that ∠APB = 2∠AQB and there is only one such point Q?

 Is there a point P on the parabola's axis of symmetry such that ∠APB = 2∠AQB and there is only one such point P?

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616947616 ● 3 years ago

point Q,no mistake in the title,Point Q is a point on the straight line y = -x-4.and point P on the parabola's axis of symmetry.

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mathmagic ● 3 years ago

mi pregunta es si es el punto P el que es unico para cada Q.

es decir el punto Q es cualquier punto del eje y el punto P es unico (en realidad hay dos considerando AQB<180º) para cada Q

supón t=y(Q) entonces y(P)=(2t^2+4t+3)/(2t)

Files: foro.ggb

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616947616 ● 3 years ago

The meaning of the question is that point P is unique. There is no limit to whether point P is unique, as long as the point P is on the axis of symmetry, the point Q is on the line y = -x-4, and the two angles mentioned in the title are two Double the relationship.

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Seror ● 3 years ago

There is indeed a P point on the parabola's axis of symmetry, and even an infinity by moving the Q point on the right y-x-4,

which verifies the relationship 'APB' to 2'AQB.

Let's trace the mediator of the AQ segment, which cuts the axis of symmetry to a P point.

Let's build the P center circle through the Q, A and B points, the APB and AQB angles

check the relationship with ABP and AQB angles (because ABP is the angle in the centre and AQB underlies the same AB arc).

The coordinates of point P are that of the center of the circle.

Files: Parabole 04.ggb

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616947616 ● 3 years ago

You may have got the parabola formula wrong. The parabola is y = - x ^ 2-4x-3, not y = - x ^ 2-4x + 2

As you did, circle P has at least two intersections with line y = - x-4.

Files: 11.ggb

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Patrick Clément ● 3 years ago

Hi,

...

Files: forumParabole2....

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616947616 ● 3 years ago

Your diagram should be correct and meet the requirements of all topics, but how can the coordinates of point P be calculated? Thank you

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mathmagic ● 3 years ago

en este diagrama Q es unico para P si distancia(P,B)=distancia(P,g)

es fácil entonces averiguar P

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616947616 ● 3 years ago

The graph should be correct and meet the requirements of the problem. So how to calculate the coordinates of P? Or help me think about how to calculate the coordinates of point P, thank you

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mathmagic ● 3 years ago

write and solve the bold equation in my post

distance(P,B)=distance(P,Q) P=(-2,y) Q=(-t-4,t) then y=(2t^2+4t+3)/(2t)

see my above post

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616947616 ● 3 years ago

Thank you, but I would like to ask how did the function "y = (2t ^ 2 + 4t + 3) / (2t)" come from in your answer? This function is not in the title?

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mathmagic ● 3 years ago

distance((x0,y0),(x1,y1))=sqrt((x1-x0)^2+(y1-y0)^2) then repeat:

write and solve the bold equation in my post

distance(P,B)=distance(P,Q) P=(-2,y) Q=(-t-4,t) then y=(2t^2+4t+3)/(2t)

https://help.geogebra.org/t...

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Seror ● 3 years ago

Just for the fun ...

The coordinates of the 3 points are :

P(-2,y) , B(-1,0), Q(-t-4,t)

Distance PQ = sqrt(-2-(-t-4))^2 + (y-t)^2) = sqrt((t+2)^2 + (y-t)^2)

Distance PB = sqrt((-2+1)^2 + (y-0)^2)=sqrt(y^2+1)

Distance PQ = Distance PB ==> Solve sqrt(y^2+1) = sqrt((t+2)^2 + (y-t)^2)

y^2+1 = ((t+2)^2 + (y-t)^2

y^2+1 = t^2+2t+4+y^2-2yt+t^2

Then ==> y=(2t^2+2t+3)/2t

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