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A question for FP1 experts

Question 6)c).
http://www.dchs-sixthform.org.uk/minisites/DCHS_mathspapers/images/FP1/June%202009/Further_Pure_1_Question_Paper%20jun%2009.pdf

I am new to FP1 and I have no idea how to figure this out, the mark scheme suggests completing the square.

Mark Scheme:
http://www.dchs-sixthform.org.uk/minisites/DCHS_mathspapers/images/FP1/June%202009/Further_Pure_1_Mark_Scheme%20jun%2009.pdf

Any help is appreciated and if possible, can you explain how to do it in a way that a stupid person like me would understand. :smile:

Thanks!
Reply 1
Avatar for aznkid66
aznkid66
1
You literally just complete the square.

4x2−8x+3y2+6y=54x^2-8x +3y^2 +6y =5

4(x2−2x+c)+3(y2+2y+d)=5+4c+3d4(x^2-2x+c) +3(y^2 +2y+d) =5+4c+3d

Find c and d such that you complete the square for x and y, respectively.

Does it make sense so far, and can you continue from there?
Reply 2
Avatar for Konnichiwa
Konnichiwa
OP
5
I don't understand how you got:

4(x2−2x+c)+3(y2+2y+d)=5+4c+3d4(x^2-2x+c) +3(y^2 +2y+d) =5+4c+3d
Reply 3
Avatar for aznkid66
aznkid66
1
Well, it's the first two steps to completing the square... Step 1. Make the leading coefficient 1 through factorization or division; Step 2. Add a constant term to complete the square.

Note that 4x2−8x+3y2+6y4x^2-8x+3y^2+6y can also be written as 4(x2)−4(2x)+3(y2)+3(2y)4(x^2)-4(2x)+3(y^2)+3(2y).

To complete the square for x, you will have to add a constant 4(c)4(c) to both sides of the equation such that x2−2x+cx^2-2x+c is a perfect square.
Similarly, to complete the square for y, you will have to add a constant 3(d)3(d) to both sides of the equation such that y2+2y+dy^2+2y+d is a perfect square.

c and d are just arbitrary constants, try not to let them confuse you XD
Reply 4
Avatar for Konnichiwa
Konnichiwa
OP
5
Oh I get it, so the value of c is -1 and the value of d is also -1, done via completing the square?

But I still have no clue what to do next. :L
Reply 5
Avatar for aznkid66
aznkid66
1
Also, note that the markscheme hints at an alternative solution that may be easier to remember/use for you.

A translation along the vector (a,0) is equivalent to a substitution of x<-x-a. A translation along the vector (0,b) is equivalent to a substitution of y<-y-b. Thus, the equation of the ellipse translated by (a,b) has the equation:

(x−a)23+(y−b)24=1\dfrac{(x-a)^2}{3} + \dfrac{(y-b)^2}{4} = 1

After expansion, and eliminating denominators by multiplying by 12, you can solve for a and b by comparison.
Reply 6
Avatar for aznkid66
aznkid66
1
Original post by Konnichiwa
Oh I get it, so the value of c is -1 and the value of d is also -1, done via completing the square?

But I still have no clue what to do next. :L


Well, no, c and d should both by +1 (when you complete the square, the constant should never be negative anyways!)...

Then you arrive at 4(x−1)2+3(y+1)2=124(x-1)^2+3(y+1)^2=12, and you divide both sides by 12 to reach the equation of the original ellipse EXCEPT with a (x-1) instead of an x and a (y+1) instead of a y, which leads you to conclude the translation was (1,-1).
Reply 7
Avatar for Konnichiwa
Konnichiwa
OP
5
Thanks, I have finally finished it. I would thumb you up but it won't let me "Please rate some other members before rating this member again.".

I owe you one!

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