FP1 Loci help

http://imgur.com/l2X5X55

Need help with 5i.
I think I got the first bit which is a circle centre (-2,2) btw is it that or do you put (-2,2i).
The line that starts at -2 has angle of 5/6pi? Is that from the x-axis so is it negative gradient?

ii. Im not sure how to work out the intersections? Im guessing its (-2, y) Not sure how to get y.

Cheers

Reply 1

Xsk

4

Original post by Super199

http://imgur.com/l2X5X55

Need help with 5i.
I think I got the first bit which is a circle centre (-2,2) btw is it that or do you put (-2,2i).
The line that starts at -2 has angle of 5/6pi? Is that from the x-axis so is it negative gradient?

ii. Im not sure how to work out the intersections? Im guessing its (-2, y) Not sure how to get y.

Cheers

Pretty sure the centre isn't (-2,2)? :tongue:

I z-x I = r represents a circle, where x is the centre and r is the radius of the circle.

The second loci is indeed a half-line starting at (-2,0), at an angle of 5pi/6 to the horizontal.

For the intersections, the way that I'd go about it is to find the equation of the circle (easy, since you know the centre and the radius) and the equation of the half-line (also easy if you say that tan(5pi/6) is the gradient and work from there) and then continue with substitution. You'll end up with two answers I think, but the value of x can't be greater than -2 or w/e because it's a half-line.
There might be a simpler way to go about it, but I can never pick up on those :tongue:

Reply 2

Super199

OP

20

Original post by Xsk

Pretty sure the centre isn't (-2,2)? :tongue:

I z-x I = r represents a circle, where x is the centre and r is the radius of the circle.

The second loci is indeed a half-line starting at (-2,0), at an angle of 5pi/6 to the horizontal.

For the intersections, the way that I'd go about it is to find the equation of the circle (easy, since you know the centre and the radius) and the equation of the half-line (also easy if you say that tan(5pi/6) is the gradient and work from there) and then continue with substitution. You'll end up with two answers I think, but the value of x can't be greater than -2 or w/e because it's a half-line.
There might be a simpler way to go about it, but I can never pick up on those :tongue:

oops my bad.
Where am I messing up

(x+2)^2+y^2= 4 , y=-root3/3 x -> y^2 = 1/3x^2

x^2+4x+4 + 1/3x^2 -4 = 0
4/3x^2 +4x =0
4x(x/3 +1) = 0

yh doesn't seem right lol.

Reply 3

Xsk

4

Original post by Super199

oops my bad.
Where am I messing up

(x+2)^2+y^2= 4 , y=-root3/3 x -> y^2 = 1/3x^2

x^2+4x+4 + 1/3x^2 -4 = 0
4/3x^2 +4x =0
4x(x/3 +1) = 0

yh doesn't seem right lol.

y=-xroot3/3 + c
c does not equal zero in this case

Btw if in doubt, draw it out. Drawing might seem silly but it can really help with visualisation

Reply 4

Super199

OP

20

Original post by Xsk

y=-xroot3/3 + c
c does not equal zero in this case

Btw if in doubt, draw it out. Drawing might seem silly but it can really help with visualisation

I think Im being daft but I can't see where it crosses the y-axis :frown:

Reply 5

Xsk

4

Original post by Super199

I think Im being daft but I can't see where it crosses the y-axis :frown:

youve worked out the gradient, and you know that it passes through the point (-2,0), therefore you can use y=mx+c to work it out.
Technically this half-line shouldnt have a y-intercept, from what I can visualise, correct? But for the sake of finding the intercept, you can treat it as a normal line until you're determining the final answer.

Reply 6

Xsk

4

If you think that's too long-winded for a mere 2 marks, there is a different method you might like more? It's harder to explain without a diagram but I can try.
When you sketch it out, you can make a triangle from (-2,0), the point of intersection, and the point on the x-axis directly below the intersection. radius=hypotenuse. You can work out that the angle in the bottom right is pi/6, use trig to work out the length of the different sides and simply work out the location of the point using these lengths and the point (-2,0)