New Senior Mathematics (Fitzpatrick) - 24(c), question 19 (1 Viewer)

[Delirium]

How do I solve this question on the parametric equation of the parabola from the Fitzpatrick Mathematics Extension 1 book:

24(c): 19. The normals to the parabola x^2 = 4ay at the point P(2at, at^2) and Q(2as, as^2) meet at R. Find the co-ordinates of R in terms of t and s. If st = -2, find the cartesian equation of the locus of R.

The back of the book has the solution: x^2 = 4a(y - 4a) -- it has no solution in regards to the co-ordinates of R.

I have attempted this question, and the troublesome area was when it came to cancelling the parameter to find the locus of R.

Any help would be greatly appreciated.

 

[B35tY]

Well if st = -2, neither s or t can be zero, can they?

 

[zeek]

Equation of Parabola -------> x²=4ay

To find the gradient of any point on the parabola you have to find the derivate .: dy/dx = m_t = x/2a

Normal at P(2at, at²)
------------------------------------------
The normal's gradient is given by m_n=-1/m_t
.: at point P the gradient of the normal is -1/t
.: Normal at P is y - y₁ = m_n(x-x₁)
.: yt=at³ + 2at - x

Normal at S(2as, as²)
-------------------------------------------
By using a similar approach as above... the normal at S becomes
ys=as³ + 2as - x

Point R
-------------
For them to meet at point R the two normals must intersect .: this parametric co ordinate can be resolved by solving the two equations simultaneously...
y_s =as² + 2a - (x/s)
y_t =at² + 2a - (x/t)

as² + 2a - (x/s) = at² + 2a - (x/t)
as² - (x/t) = as² - (x/s)
.: as² - at² = x((1/s)-(1/t))
= x((t-s)/st))
.:[a(s² - t²).st]/(t-s) = x <=== factorise numerator
.:x = -ast(t+s) <======== re-substitute to get y
.:y =at² + 2a + as(t+s)

Cartesian equation of R
--------------------------------
To find the cartesian equation of R, remove the parametrics so...
st=-2

x=-ats(t+s)
=2a(t+s)
.: x/2a = t+s

y=at² + 2a + as(t+s)
=a(t² + ts + s² +2)
=a(t² + s²)
=a(t+s)² -2ats
=a(t+s)² + 4a <====== substitute for x

y=a(x/2a)² + 4a
=x² + 4a
4ay=x² + 16a²
.:4a(y-4a)=x²

EDIT: lol i didn't understand what you meant in your comment :S but yes you can assume certain things provided you give a substantial explanation or proof, however, this is usually done when you are actually working through the question :S
Keep thinking outside of the box though... maybe that way you can make it to 4 unit

 

[Delirium]

Thank you very much. Both of you have helped me. Stupid old me slightly misinterpreted the question, and also didn't know what to do in some respects. I especially like the way the locus was obtained in the end (by the elimination of s + t), I never would've thought to do it like that -- I was especially curious about this, so that itch has been scratched so to speak.

Again, thank you.

 

[Delirium]

Sorry to bump this thread up, but there's one place of uncertainty regarding r3v3ng3's solution... In the section of his solution where he is finding the point R: when he is finding the y-coordinate of this point he substituted the x-coordinate which he found into one of the equations he began with.

The coordinate he gives is:

y = at² + 2a + as(t+s)

But if you substitute the value of x into the original equation he used yourself, you will end up with this:

y = at² + 2a + ast(t+s)/t

Obviously, the variable t has been cancelled. But the question has not told you to use the condition st = -2 in finding the coordinates of R. So, it's my opinion that in the solution for finding R, the condition "when t does not equal zero" should be included, since if st = -2 is not given as a condition, it is perfectly valid to assume that t can equal zero and that there still is an intersection point, yet its y-coordinate may not be that which is stated [i.e. y = at² + 2a + as(t+s)], as a division of the variable t has been performed, which in the case of t = 0 would be invalid.

Am I right? Or is there some other reasoning to justify his solution (which is otherwise correct)?

 

[Riviet]

If you look back at the equation of the normal to P:

y =at² + 2a - (x/t)

t can't equal zero because this would result in an undefined denominator in x/t.

Hope that clears things up.

 

[Delirium]

@Riviet: True, but the fact is that a normal does exist at the point t = 0, with its equation being x = 0 (i.e. the axis of the parabola). Therefore, I would say that it would be rightful to exclude t = 0 from the given coordinates of R, as the methods used to obtain R assume that t does not equal zero...

 

[Riviet]

@Riviet: True, but the fact is that a normal does exist at the point t = 0, with its equation being x = 0 (i.e. the axis of the parabola). Therefore, I would say that it would be rightful to exclude t = 0 from the given coordinates of R, as the methods used to obtain R assume that t does not equal zero...

I see what you mean, then I guess a note next to that line (after dividing by t) would be better, ie "t =/= 0".