Parametric Equations/ Range and Domain

Original post by SapphirePhoenix1

Hey. Can anyone help with this exam question?

The curve C has parametric equations
x= 3 + 2sqrt(3) cos t, y = 5sqrt(3) + 2sqrt(3) sin t, -pi/4 <= t <= 2pi/3

a) Show that all points on C satisfy (x-3)^2 + (y-5sqrt(3))^2 = 12 (4)
DONE

b) For curve C, (i) state the range of x, (ii) state the range of y (2)
DONE

The point P lies on C.
Given the line with equation y = mx + 12sqrt(3), where m is a constant, intersects C at P,
c) state the range of m, writing your answer in set notation
NEED HELP
So far, I've implicitly differentiated the equation for C and got dy/dx = (3-x)/(y-5sqrt(3)), but when I plug in any of the values I got in b I don't get any of the correct answers (my teacher gave us the numerical answers)

The points (0,0), (0,12sqrt(3)) and P form a triangle.
d) (i) Find the largest possible area of the triangle
d (ii) Find the smallest possible area of the triangle
Can't do this without having done part c.

If anyone can offer any guidance, I'd really appreciate it.

Thanks in advance :smile:

I'm sure you realise it's the equation of a circle? So you want to find where the line/circle intersect. You could do it using discriminant, tangent geometry, ... if the range of t is not important Have you sketched it?
Obviously you need to take the range of t into account if it is important. The x-y end points would "give" the gradients

(edited 4 years ago)

Reply 2

OP

Original post by mqb2766

I'm sure you realise it's the equation of a circle? So you want to find where the line/circle intersect. You could do it using discriminant, tangent geometry, ... if the range of t is not important Have you sketched it?
Obviously you need to take the range of t into account if it is important. The x-y end points would "give" the gradients

ok, I got one of answers to part c

You were meant to plug in the ranges into the parametric equations (as you said) but then use the fact that the y-intercept is (0,12rt(3)) (from y = mx + 12rt(3)) and work out the gradient from that, i.e. ((12rt(3)-(5rt(3)+2rt(3)sin(2pi/3))/((0-(3+2rt(3)cos(2pi/3)))

but I still can't get the other one

(edited 4 years ago)

Reply 3

Original post by SapphirePhoenix1

ok, I got one of answers to part c

You were meant to plug in the ranges into the parametric equations (as you said) but then use the fact that the y-intercept is (0,12rt(3)) (from y = mx + 12rt(3)) and work out the gradient from that, i.e. ((12rt(3)-(5rt(3)+2rt(3)sin(2pi/3))/((0-(3+2rt(3)cos(2pi/3)))

but I still can't get the other one

Have you sketched it? If so upload your sketch and working.

Reply 4

OP

The sketch is in the question

20201206_123337-compressed.jpg.jpeg

Screenshot_20201206-123312_OneNote-compressed.jpg.jpeg

Reply 5

OP

the -2(r3) is correct, but I don't think you were supposed to get it like that XD

Reply 6

Original post by SapphirePhoenix1

the -2(r3) is correct, but I don't think you were supposed to get it like that XD

The top point is obviously the steepest gradient (most negative).
The other end of the range of m (least steep or negative) must be when the line forms a tangent to the circle. Just sketch it on there. So use the discriminant/geometry to solve it.

Reply 7

OP

I used Desmos earlier to sketch the graph and the lines with the upper and lower values of m (got them from the numerical answers).
The red dot and blue dot show where the graph starts and ends

workings01.png217.5KB

Reply 8

OP

Original post by mqb2766

The top point is obviously the steepest gradient (most negative).
The other end of the range of m (least steep or negative) must be when the line forms a tangent to the circle. Just sketch it on there. So use the discriminant/geometry to solve it.

I don't quite understand how you could use the discriminant for this

Reply 9

Original post by SapphirePhoenix1

I don't quite understand how you could use the discriminant for this

It's a gcse technique? Sub the equation of the line (solve them simultaneously) into the circle and work out where the resulting quadratic has a single solutuon, or the discriminant is zero.

Reply 10

OP

Original post by mqb2766

It's a gcse technique? Sub the equation of the line (solve them simultaneously) into the circle and work out where the resulting quadratic has a single solutuon, or the discriminant is zero.

Well, I am a gcse (add maths) student, so I don't know what you hoped to gain from that. My teacher literally gave this to me and told me to try it.

Anyway...
I subbed y = mx +12rt(3) into (x-3)^2 + (y-5rt(3))^2 = 12 and got
(x-3)^2 + (mx+7rt(3)^2 = 12, so I don't really see how you could use the discriminant for that... (unless I subbed it into the wrong equation...)

Reply 11

Original post by SapphirePhoenix1

Well, I am a gcse (add maths) student, so I don't know what you hoped to gain from that. My teacher literally gave this to me and told me to try it.

Anyway...
I subbed y = mx +12rt(3) into (x-3)^2 + (y-5rt(3))^2 = 12 and got
(x-3)^2 + (mx+7rt(3)^2 = 12, so I don't really see how you could use the discriminant for that... (unless I subbed it into the wrong equation...)

Parametric curves and circles not centred on the origin are both beyond standard gcse. The discriminant technique is standard gcse. I hoped to gain nothing apart from being clear you should know it if you're attempting this question.

You have a quadratic in x. When there is a single solution the line is a tangent. This occurs when the discriminant (which depends on m) is zero.
https://www.bbc.co.uk/bitesize/guides/z9pssbk/revision/5

(edited 4 years ago)

Reply 12

OP