minding the ps and qs

**s_ingram** · July 28th 2009, 03:12 AM

Hi guys!

I am baffled by my own solution to the following parabola question. Having got the right answer, I find that I don't like it! Can anyone help?

The parabola is $y^2 = 4ax$ and at point $P(ap^2, 2ap)$ the tangent is found to be $yp = x + ap^2$ . We have to find the coordinates of the point of intersection T of the tangent at P and the tangent at $Q(ap^2, 2aq)$ .

In the derivation of the tangent at P the gradient is found to be 1/p.
To get the tangent at Q, given that P and Q have the same coordinate $ap^2$ , I would expect the intersection to be on the x-axis (since $y^2 = 4ax$ is symmetrical). The answer does not confirm this, so that's my first question - why not?

If I put the coords of Q into the parabola I get:

$(2aq)^2 = 4a(ap^2)$
$q^2 = p^2$
$q = \pm p$

I therefore take the gradient at Q to be -1/p, then the equation of the tangent follows:

$\frac{y - 2aq)}{x - ap^2} = \frac{-1}{p}$
call this equation 1
$yp - 2apq = -x + ap^2$
The tangent at P is
$yp = x + ap^2$
subtracting gives
$x = apq$
$y = a(p + q)$
which is the correct answer.

The thing that puzzles me is that to get the answer I am assuming the gradient at Q is symmetrical i.e. -1/p (since the gradient at P is 1/p). Also if I express this gradient as 1/q which is the same thing and substitute into equation 1, I get:

$\frac{y - 2aq)}{x - ap^2} = \frac{1}{q}$
$yq - 2aq^2 = x - ap^2$
and equation 1 is:
$yp = x + ap^2$
subtracting:
$y(p - q) - 2aq^2 = -2ap^2$
$y(p - q) = 2a(q + p)(q - p)$
$y = 2a(q + p)$

A different result! All these ps and qs are swimming before my eyes. Have I made a slip somewhere?

**HallsofIvy** · July 28th 2009, 04:16 AM

Originally Posted by s_ingram

Hi guys!

I am baffled by my own solution to the following parabola question. Having got the right answer, I find that I don't like it! Can anyone help?

The parabola is $y^2 = 4ax$ and at point $P(ap^2, 2ap)$ the tangent is found to be $yp = x + ap^2$ . We have to find the coordinates of the point of intersection T of the tangent at P and the tangent at $Q(ap^2, 2aq)$ .

I presume that q is not p but then q= -p. So Q is the point $(ap^2, -2ap)$ .

In the derivation of the tangent at P the gradient is found to be 1/p.
To get the tangent at Q, given that P and Q have the same coordinate $ap^2$ , I would expect the intersection to be on the x-axis (since $y^2 = 4ax$ is symmetrical). The answer does not confirm this, so that's my first question - why not?

If I am understanding what you wrote correctly, yes, the point of intersection of the tangents at $P(ap^2, 2ap)$ and $Q(ap^2, -2ap)$ intersect on the x-axis.

If I put the coords of Q into the parabola I get:

$(2aq)^2 = 4a(ap^2)$
$q^2 = p^2$
$q = \pm p$

I therefore take the gradient at Q to be -1/p, then the equation of the tangent follows:

$\frac{y - 2aq)}{x - ap^2} = \frac{-1}{p}$
call this equation 1
$yp - 2apq = -x + ap^2$

Okay. But q= -p so this is the same as $yp- 2ap^2= -x+ ap^2$

The tangent at P is
$yp = x + ap^2$
subtracting gives
$x = apq$
$y = a(p + q)$
which is the correct answer.

And, of course, q= -p so this is $x= ap^2$ and $y= 0$

The thing that puzzles me is that to get the answer I am assuming the gradient at Q is symmetrical i.e. -1/p (since the gradient at P is 1/p). Also if I express this gradient as 1/q which is the same thing and substitute into equation 1, I get:

$\frac{y - 2aq)}{x - ap^2} = \frac{1}{q}$
$yq - 2aq^2 = x - ap^2$
and equation 1 is:
$yp = x + ap^2$
subtracting:
$y(p - q) - 2aq^2 = -2ap^2$
$y(p - q) = 2a(q + p)(q - p)$
$y = 2a(q + p)$

A different result! All these ps and qs are swimming before my eyes. Have I made a slip somewhere?

No, it's not a different result. q is still -p, q+ p= 0, and 2a(0)= a(0)= 0. The only "slip" you made was not to replace q by -p as soon as you realized, as you say above, that $q= \pm p$ . Of course, q= p does not give two points so we don't have two tangent to intersect and the whole question falls apart: q= -p.

**s_ingram** · July 28th 2009, 05:28 AM

You have unconfused me! Thank you.

**Opalg** · July 28th 2009, 08:24 AM

Originally Posted by s_ingram

Hi guys!

I am baffled by my own solution to the following parabola question. Having got the right answer, I find that I don't like it! Can anyone help?

The parabola is $y^2 = 4ax$ and at point $P(ap^2, 2ap)$ the tangent is found to be $yp = x + ap^2$ . We have to find the coordinates of the point of intersection T of the tangent at P and the tangent at $Q(ap^2, 2aq)$ .

Are you sure that you have copied the question correctly? It seems more likely to me that the point Q should be $(a{\color{red}q}^2, 2aq)$ . Then the tangent at P has equation $yp = x + ap^2$ and the tangent at Q has equation $yq = x + aq^2$ . Subtract the second of those equations from the first, divide by p–q, and you get the answer $y=a(p+q)$ .

**HallsofIvy** · July 28th 2009, 01:13 PM

Originally Posted by Opalg

Are you sure that you have copied the question correctly? It seems more likely to me that the point Q should be $(a{\color{red}q}^2, 2aq)$ . Then the tangent at P has equation $yp = x + ap^2$ and the tangent at Q has equation $yq = x + aq^2$ . Subtract the second of those equations from the first, divide by p–q, and you get the answer $y=a(p+q)$ .

No, the problem is clearly for the two points where a vertical line crosses the parabola. Otherwise the whole problem makes no sense.

**Opalg** · July 29th 2009, 12:22 AM

Originally Posted by HallsofIvy

No, the problem is clearly for the two points where a vertical line crosses the parabola. Otherwise the whole problem makes no sense.

On the contrary, the problem makes perfect sense if $P=(ap^2,2ap)$ and $Q=(aq^2,2aq)$ . The tangents at these two general points on the parabola meet at the point $(apq,a(p+q))$ , whose y-coordinate is a(p+q), as required.

If the problem refers to two points where a vertical line crosses the parabola, then we would have to have q=–p, and the tangents would intersect on the x-axis. But the answer indicates that this is not the situation, as the OP noted:

Originally Posted by s_ingram

To get the tangent at Q, given that P and Q have the same coordinate $ap^2$ , I would expect the intersection to be on the x-axis (since $y^2 = 4ax$ is symmetrical). The answer does not confirm this, so that's my first question - why not?

**HallsofIvy** · July 29th 2009, 05:44 AM

S_ingram didn't say what "answer" he was referring to. I thought he meant the answer he had got to the problem: y= p+ q. Of course, your interpretation is reasonable- assuming that he had copied the problem incorrectly.

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