Yet another parametrics question (1 Viewer)

[5647382910]

this question is really pissing me off, any help would be much appreciated:

Through the vertex A of a parabola, chords AP and AQ are drawn at right angles to one another. Show that for all positions of P, PQ cuts the axis of the parabola in a fixed point K

What i did was assume the parabola was in the form (x - h)^2 = 4a(y - k) and so i represented P by (2at + h, at^2 + k) and gave the other points some pronumerals but from then on everything is messed.

 

[Michaelmoo]

Sometimes these parametrics may look tricky, but the good thing about this topic is that it is repetition of the same logic.

First we establish the relationship between parameters.

As you said, let the vertex = A(h,k) and P = (2ap + h, ap^2 + k) and similarly Q = (2aq + h, aq^2 + k).

Now the relationship is that the lines AP and AQ are perpendicular (right angles), therefore the multiplication of their gradients is -1. We use the gradient formula to find gradients:

Gradient of AP x Gradient of AQ = -1

(ap^2 +k) - k'''''''''''''''(aq^2 + h) - h
--------------- '''''''''x''' -------------------
(2ap + h) - h''''''''''''''''(2aq + h) - h

= -1

Now Simplify this expression, Factorise and cancel out, You will get pq = -4

Now we have our linking expression we go to find the equation of the chord PQ:

'''''''''''(ap^2 +k) - (aq^2 + k)
m = -----------------------------
'''''''''''(2ap + h) = (2aq + h)


= (p+q)
-----------
......2


You'll find that after you simplify, the gradient of your chord PQ = (p+q)/2

Now to Find the equation of the chord PQ, use the gradient you've just found and point P (point gradient formula)

y- (ap^2 +k) = (p+q) x [x-(2ap + h)]
'''''''''''''''''''''''''''''''--------
'''''''''''''''''''''''''''''''''''''2


Simplify this and you get your Equation of your chord to be:

2y - 2ap^2 - 2k = (p+q)x - (p+q)(2ap + h)

Now the final step is to answer the question, find where this point intersects the axis of the parabola. In this case the axis is x = h. Sub it in to the equation of the Chord PQ and simplify!!!

2y - 2ap^2 - 2k = (p+q)h - (p+q)(2ap + h)

2y - 2ap^2 - 2k = ph + qh -2ap^2 -ph -2apq - 2h

2y = 2k -2apq

Now this is where our relationship (pq = -4) comes in hand we sub this into the equation to eliminate variables P and Q.

2y = 2k + 8a

y = k + 4a

This means that under these conditions, the chord will ALWAYS cut the axis of the parabola at (h, k+4a), and since h, k and a are constants to the parabla, Therefore it will always cut at a fixed point k.

What this means is basically for whatever points you choose P and Q to be, it always cuts at these points because the vertex of that particula parabola (h,k) is always the same and so is the focal length (a). i.e. the point is only dependant on these 2 constant factors, not on the VARIABLE parameters (p, q).

Hope that helps and good luck!!!

 

[5647382910]

Sometimes these parametrics may look tricky, but the good thing about this topic is that it is repetition of the same logic.

First we establish the relationship between parameters.

As you said, let the vertex = A(h,k) and P = (2ap + h, ap^2 + k) and similarly Q = (2aq + h, aq^2 + k).

Now the relationship is that the lines AP and AQ are perpendicular (right angles), therefore the multiplication of their gradients is -1. We use the gradient formula to find gradients:

Gradient of AP x Gradient of AQ = -1

(ap^2 +k) - k'''''''''''''''(aq^2 + h) - h
--------------- '''''''''x''' -------------------
(2ap + h) - h''''''''''''''''(2aq + h) - h

= -1

Now Simplify this expression, Factorise and cancel out, You will get pq = -4

Now we have our linking expression we go to find the equation of the chord PQ:

'''''''''''(ap^2 +k) - (aq^2 + k)
m = -----------------------------
'''''''''''(2ap + h) = (2aq + h)


= (p+q)
-----------
......2


You'll find that after you simplify, the gradient of your chord PQ = (p+q)/2

Now to Find the equation of the chord PQ, use the gradient you've just found and point P (point gradient formula)

y- (ap^2 +k) = (p+q) x [x-(2ap + h)]
'''''''''''''''''''''''''''''''--------
'''''''''''''''''''''''''''''''''''''2


Simplify this and you get your Equation of your chord to be:

2y - 2ap^2 - 2k = (p+q)x - (p+q)(2ap + h)

Now the final step is to answer the question, find where this point intersects the axis of the parabola. In this case the axis is x = h. Sub it in to the equation of the Chord PQ and simplify!!!

2y - 2ap^2 - 2k = (p+q)h - (p+q)(2ap + h)

2y - 2ap^2 - 2k = ph + qh -2ap^2 -ph -2apq - 2h

2y = 2k -2apq

Now this is where our relationship (pq = -4) comes in hand we sub this into the equation to eliminate variables P and Q.

2y = 2k + 8a

y = k + 4a

This means that under these conditions, the chord will ALWAYS cut the axis of the parabola at (h, k+4a), and since h, k and a are constants to the parabla, Therefore it will always cut at a fixed point k.

What this means is basically for whatever points you choose P and Q to be, it always cuts at these points because the vertex of that particula parabola (h,k) is always the same and so is the focal length (a). i.e. the point is only dependant on these 2 constant factors, not on the VARIABLE parameters (p, q).

Hope that helps and good luck!!!

dude, u are the best
thanks heaps

 

[5647382910]

just one query... the part which screwed me was the fact that point Q was fixed so i didnt write it as a parametric equation, does ur solution still apply?