Yet another locus question (1 Viewer)

[z600]

a parabola with the vertext (-1,-1) meets xy=1 again at A and B, find M (mid point of of A and B) and hence the locus.

I used y=a(x-h)^2+k

i subbed in (-1, -1) and found the two roots and the mid point to be 3? then its just a mess from then onwards.

ALSO

WHen a polynomial is divided by (2x+1)(x-3), the remainder is 3x-1, What is the remainder when the polynomial is divided by 2x+1?

 

[Evergreen]

What is "find M of A and B" supposed to mean?

For the second one:
Let P(x) be the unknown polynomial
when we divide this polynomial we get a quotient Q(x) plus remainder 3x - 1, so:
P(x) / [(2x + 1)(x - 3)] = Q(x) + {3x - 1}
NB: the thing in {} is the remainder

now what happens when we divide by 2x + 1 only? mulitply both sides by (x - 3)
P(x) / (2x + 1) = (x - 3)Q(x) + {(3x - 1)(x - 3)}
Remember that the thing in the thing in {} is the remainder as (x - 3)Q(x) is the new quotient

.: Remainder is (3x - 1)(x - 3)

I think that seems right....

Thats a pretty neat way to do it and I think he is talking about the midpoint of the line AB.

 

[z600]

I did it that way at first but its wrong...the answer is 1/2

I got -2.5 as the answer and my teacher agrees with me

 

[Chinmoku03]

a parabola with the vertext (-1,-1) meets xy=1 again at A and B, find M (mid point of of A and B) and hence the locus.

I used y=a(x-h)^2+k

i subbed in (-1, -1) and found the two roots and the mid point to be 3? then its just a mess from then onwards.

Ok, if the vertex of a parabola is (-1, -1), then its equation should be y = (x+1)²-1. Since xy = 1, y = 1/x. Equate the two equations to get 1/x = x²+2x, which can be rearranged to x²+2x- (1/x) = 0. Multiply both sides by x to give x³+2x²-1 = 0.

Now, since x = -1 is a root of this equation, x³+2x²-1 can be written as (x+1)(x²+ax-1). Comparing the coefficients of x², 2 = 1+a, therefore a = 1. So x³+2x²-1 can then be written as (x+1)(x²+x-1). Since x³+2x²-1 = 0, then (x+1)(x²+x-1) = 0. Using the quadratic formula, x₁ = (-1+sqrt5)/2 and x₂ = (-1-sqrt5)/2, where sqrt means square root.

Sub x₁ and x₂ back into either the equation of the parabola or hyperbola to find the value of y₁ and y₂ respectively. After that, finding the midpoint's easy, just rather tedious. M = (-1/2, -1/2) from my working out.

Sorry, but I don't remember how to work out the locus anymore unless it's parametric >.<; Does anybody else have an idea?

 

[Trebla]

Oh Crap, I just realised I did it wrong...ignore what I said earlier....lol
Here's the correct answer (I hope!)
Let P(x) be the unknown polynomial
when we divide this polynomial we get a quotient Q(x) plus remainder 3x - 1, so:
P(x) / [(2x + 1)(x - 3)] = Q(x) + (3x - 1) / [(2x + 1)(x - 3)]
i.e. P(x) = (2x + 1)(x - 3)Q(x) + 3x - 1

now when we divide by (2x + 1), we can use the remainder theorem:
P(-1/2) = 0 - 3/2 - 1
Hence the remainder is - 5/2

 

[Trebla]

The equation of the parabola is much more general than that. It can scale according to multiples of its coefficients.

Let the equation of the parabola be y = a(x + 1)² - 1
To find points of intersection:
1/x = a(x + 1)² - 1
.: ax³ + 2ax² + x(a - 1) - 1 = 0
We know that one of the points of intersection and hence a root of this equation is x = - 1. Let the other two roots be x1 and x2.
Sum of roots: x1 + x2 - 1 = - 2
.: (x1 + x2)/2 = -1/2

Now solve for y coordinates:
y = a(1/y + 1)² - 1
.: y³ - y²(a - 1) - 2ay - a = 0
We know that one of the points of intersection and hence a root of this equation is y = - 1. Let the other two roots be y1 and y2.
Sum of roots: y1 + y2 - 1 = a - 1
.: (y1 + y2)/2 = a/2

.: M (-1/2 , a/2)

The locus of M is pretty obvious once you found it. Since 'a' can vary for any non-zero real number, the locus should simply be x = -1/2

 

[onebytwo]

hey z600, please make another post asap, seeing 666 is unsettling.

 

[z600]

hey z600, please make another post asap, seeing 666 is unsettling.

LOL thanks guys

For the Polynimial one, i got -2.5 as well but the answer says 0.5
And apparantly, when you shove in a polynomial function both 0.5 or -2.5 doesnt work:mad1: :mad1:

 

[z600]

Solution:

View attachment 15322

Nicee thanks

The guys who wrote the question was
Dr Bill Pender, Master in Charge of Mathematics, Sydney Grammar School

 

[chousta]

Solution:

View attachment 15322

not a very nice thing to say vafa. "was absolutely a stupid", and by the way you expressed that you seem to be the "stupid" one, but hey...no one is pointing fingers...

 

[Trebla]

For the Polynimial one, i got -2.5 as well but the answer says 0.5
And apparantly, when you shove in a polynomial function both 0.5 or -2.5 doesnt work:mad1: :mad1:

Are you sure they don't work?
Consider the polynomial P(x) = 2x² - 2x - 4
If you expand (2x + 1)(x - 3) + 3x - 1, you should get that polynomial P(x), so it does give a remainder of 3x - 1 when divided by (2x + 1)(x - 3)
When you divide P(x) by (2x + 1), you get a quotient of (x - 3/2), with a remainder of - 5/2....