Parabola (1 Viewer)

[OLDMAN]

A question worthy of q7 3U, or a definite Harder 3Unit.

The chord AB is normal to the parabola x^2=4ay. Find the point A which minimizes the length of this chord.

 

[Harimau]

Originally posted by OLDMAN
A question worthy of q7 3U, or a definite Harder 3Unit.

The chord AB is normal to the parabola x^2=4ay. Find the point A which minimizes the length of this chord.

Im going to take a wild guess and say the origin... or (1,1).

 

[underthesun]

this is how far i got::

got the normal equation

ty - at^3 = 2at - x

ty + x = 2at + at^3

now, the normal meets the parabola again at
(2aq, aq^2)

we sub values in

atq^2 + 2aq = 2at + at^3

atq^2 + 2aq - 2at - at^3 = 0

this is a quadratic in q

sum of roots = -b/a (from polynomials)

q1 + q2 = -2a / at

however, we know that one of the roots is t

q2 + t = -2a/at

q2 = -(2a + at^2)/at

Now we know the q value for the intersection point

I subbed the valu in (2aq, aq^2),

Q(-4a/t - 2at, (a/t^2)(4 + 4t^2 + t^4) )

and the original point of normal

T(2at, at^2)

and i'd use the distance formula,

QT^2 = (2at - (-4a-2at))^2 + (at^2 - (a/t^2)(4+4t^2 + t^2)

which simplifies to

QT^2 = 16a^2 (1 + 1/t) (t^2 + 2 + 1/t^2)

QT^2 = 16a^2 (t^2 + t + 2 + 2/t + 1/t^2 + 1/t^3)

now.. finding the minimum..
mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

or am I stuck on the wrong direction?

 

[underthesun]

well.. I used the graph calculator, and found that for the minimum length of AB, t = 1.25 something..

irrational value. confusing question. How many marks would this be?

(and if you were a marker, could you be kind enough to tell me how much mark would I have scored )

 

[OLDMAN]

Underthesun: you've done well.

This question is much a test of algebraic manipulation. Small error, corrected as follows :

QT^2 = (2at - (-4a/t-2at))^2 + ((at^2 - (a/t^2)(4+4t^2 + t^4))^2
=(16a^2)(t+1/t)^2+a^2(4/t^2+4)^2
=(16a^2)((t+1/t)^2+(1/t^2+1)^2)
=(16a^2)((t^2+1)^2)(1/t^2+1/t^4)
=[(16a^2)((t^2+1)^3]/t^4

I'll let you finish it from here...

 

[underthesun]

ahh.. i see

QT^2 = [16a^2 (t^2 + 1)^3] / t^4

differentiate it to find stationary points of QT^2,

(my working would be a mess if I typed it hard, but you get the picture)

d/dt = 16a^2[(t^2+1)^2 / t^5] * [ 2t^2 - 4]

for d/dx = 0,

2t^2 - 4 = 0

hence,

t^2 = 2

t = (positive or negative) root two

Normally I would say this

by looking at the graph of parabola, there would be no "maximum" for QT^2 since that the distance can get infinitely big for QT^2 at infinite values of t, hence t = sqrt(2) and t=-sqrt(2) are the values of t in which the chord made by the normal from point A is the shortest. Hence the point A is (sub t in to 2at, at^2) or (the other negative one).

is that sufficient for proving that it is minimum? or do I have to differentiate or use table method for decreasing and increasing function value etc..

need to brush up my algebra skills

thanks for the question.. made me realise my holes in differentiation

 

[OLDMAN]

Quote:
by looking at the graph of parabola, there would be no "maximum" for QT^2 since...

If I was the marker yes.

However, just in case you did try to prove minima by the usual way ie. second derivative, or increasing-decreasing table, I always tell my students (2,3 and 4U) that when you have a potentially messy derivative (products, quotients) the second method (increasing/decreas) works more efficiently, particularly when terms are in convenient factorized form eg.

d/dt = 16a^2[(t^2+1)^2 / t^5] * [ 2t^2 - 4]

it is easy to see that slightly to the left of +sqrt(2) it is neg., slightly to the right, positive etc. And when it is -sqrt(2) the t^5 in the denominator makes sure that you get the same result. So there, it is a myth that the second derivative method is always a shortcut!

Good work btw.

 

[underthesun]

Dear sir,

I hope all markers are like you..

 

[OLDMAN]

underthesun quote:
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by looking at the graph of parabola, there would be no "maximum" for QT^2 since that the distance can get infinitely big for QT^2 at infinite values of t, hence t = sqrt(2) and t=-sqrt(2) are the values of t in which the chord made by the normal from point A is the shortest. Hence the point A is (sub t in to 2at, at^2) or (the other negative one).
--------------------------------------------------------------------------------

We can mathematically translate your beautiful quote to dispel all traces of marker's doubt :

the function QT^2 = [16a^2 (t^2 + 1)^3] / t^4 has vertical asymptote at t=0, it is always positive, as well it approaches +infinity as t----+-infinity. Hence, since there is but one stationary point for t>0 at t=sqrt(2), then it must be the minimum. Similarly for t<0.
The new Complex Number Q will have a still more interesting discussion in finding and determining (nature of) turning points.

 

[underthesun]

If i could write that I would write that, but i was thinking of this:

If the above question appears in an exam, and I got up to the point of proving the nature of turning point. But I already know it is a minimum, I just don't know how to prove it.

Then just say, that an idea came up to my mind. I just need to create an increasing-decreasing table but not do any calculations, and just answering as -ve or +ve (indicating decreasing or increasing), because I already know the nature of the turning point.

Would the marker be satisfied with "+ve" or "-ve" values in the increasing-decreasing table?

 

[OLDMAN]

I believe, in the context of this problem -considering that since you are in no doubt that it is a minima, yes go ahead and take a punt - I might get shot down on this.

But be careful, that one day examiners might come up with a question specifically to catch the -ve and +ve bluffers!