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Please could someone tell me how to do this question, I have the answers, just dont know how to do it.

6. A curve is given parametrically by the equations

x = 5 cos t, y = -2 + 4 sin t, 0 £ t < 2p.

(a) Find the coordinates of all the points at which C intersects the coordinate axes, giving your answers in surd form where appropriate.

(4 marks)

6. A curve is given parametrically by the equations

x = 5 cos t, y = -2 + 4 sin t, 0 £ t < 2p.

(a) Find the coordinates of all the points at which C intersects the coordinate axes, giving your answers in surd form where appropriate.

(4 marks)

x = 5 cos t, y = -2 + 4 sin t, 0 £ t < 2p.

wen x=0

5cos t = 0

cos t = 0

t = whatever

wen y=0

4sin t = 2

sint =0.5

t = pi / 6 or whatever it is

wen x=0

5cos t = 0

cos t = 0

t = whatever

wen y=0

4sin t = 2

sint =0.5

t = pi / 6 or whatever it is

BlueAngel

Please could someone tell me how to do this question, I have the answers, just dont know how to do it.

6. A curve is given parametrically by the equations

x = 5 cos t, y = -2 + 4 sin t, 0 £ t < 2p.

(a) Find the coordinates of all the points at which C intersects the coordinate axes, giving your answers in surd form where appropriate.

(4 marks)

6. A curve is given parametrically by the equations

x = 5 cos t, y = -2 + 4 sin t, 0 £ t < 2p.

(a) Find the coordinates of all the points at which C intersects the coordinate axes, giving your answers in surd form where appropriate.

(4 marks)

Set x = 0...find values for t and then corresponding values of y.

Then set y = 0, find values for t and corresponding values of x

These coordinates you get are the points wehere the curve cuts the axes.

G

BlueAngel

Please could someone tell me how to do this question, I have the answers, just dont know how to do it.

6. A curve is given parametrically by the equations

x = 5 cos t, y = -2 + 4 sin t, 0 £ t < 2p.

(a) Find the coordinates of all the points at which C intersects the coordinate axes, giving your answers in surd form where appropriate.

(4 marks)

6. A curve is given parametrically by the equations

x = 5 cos t, y = -2 + 4 sin t, 0 £ t < 2p.

(a) Find the coordinates of all the points at which C intersects the coordinate axes, giving your answers in surd form where appropriate.

(4 marks)

x=0 therefore cost = 0 => t = pi/2 or 3pi/2

y = 2 or -6 giving (0,2) and (0,-6)

y=0 => sint = 1/2 => t= pi/6 or 5pi/6

x = (root3)/2 and -(root3/2)

giving (root3/2,0) AND (-root3/2,0)

it is an ellipse

MB

rockindemon

x = 5 cos t, y = -2 + 4 sin t, 0 £ t < 2p.

wen x=0

5cos t = 0

cos t = 0

t = whatever

wen y=0

4sin t = 2

sint =0.5

t = pi / 6 or whatever it is

wen x=0

5cos t = 0

cos t = 0

t = whatever

wen y=0

4sin t = 2

sint =0.5

t = pi / 6 or whatever it is

I understand that - when x = 5 cos t t = pi/2, and 3pi/2

when y = -2 + 4 sin t t = pi/6 etc...

But thats not the answer, the answer is (0, 2) , (0, -6) , (5root3/2, 0 ) , and (-5root3/2 , 0). Which I dont understand how you get.

gzftan

Tut tut!! BlueAngel hasn't got that far yet!!!!! You're jumping ahead!! lol

G

G

sorry

MB

Oh my god, I understand now. Thank you everyone.

That question now looks so easy, Im so thick, Im gonna fail tomoz.

That question now looks so easy, Im so thick, Im gonna fail tomoz.

BlueAngel

I understand that - when x = 5 cos t t = pi/2, and 3pi/2

when y = -2 + 4 sin t t = pi/6 etc...

But thats not the answer, the answer is (0, 2) , (0, -6) , (5root3/2, 0 ) , and (-5root3/2 , 0). Which I dont understand how you get.

when y = -2 + 4 sin t t = pi/6 etc...

But thats not the answer, the answer is (0, 2) , (0, -6) , (5root3/2, 0 ) , and (-5root3/2 , 0). Which I dont understand how you get.

yep, sorry I made a silly error. x = 5cost. cos t = root3/2 and -root3/2 so x = (5root3/2, 0 ) , and (-5root3/2 , 0)

MB

not posted before so i'm not sure if my notations are correct....but....

x=5cost

y=-2+4sint

x/5=cost

y+2/4=sint

using sin^2 t+cos^2t=1

(x/5)^2 + (y+2/4)^2=1

then substitute x=0 and y=0 into this equation and you get

y=-6 or y=2

and then x=+5root3/2 or -5root3/2

hope this helps

helen x

x=5cost

y=-2+4sint

x/5=cost

y+2/4=sint

using sin^2 t+cos^2t=1

(x/5)^2 + (y+2/4)^2=1

then substitute x=0 and y=0 into this equation and you get

y=-6 or y=2

and then x=+5root3/2 or -5root3/2

hope this helps

helen x

Miss Jones

not posted before so i'm not sure if my notations are correct....but....

x=5cost

y=-2+4sint

x/5=cost

y+2/4=sint

using sin^2 t+cos^2t=1

(x/5)^2 + (y+2/4)^2=1

then substitute x=0 and y=0 into this equation and you get

y=-6 or y=2

and then x=+5root3/2 or -5root3/2

hope this helps

helen x

x=5cost

y=-2+4sint

x/5=cost

y+2/4=sint

using sin^2 t+cos^2t=1

(x/5)^2 + (y+2/4)^2=1

then substitute x=0 and y=0 into this equation and you get

y=-6 or y=2

and then x=+5root3/2 or -5root3/2

hope this helps

helen x

Hey that another method, but its a good one. Thanx.

erm sorry if im wrong or something but

i agree that the parametric gives an elipse shape

but i dont think its in terms of pi

x = 5 cos t, y = -2 + 4 sin t

the furthest distance of x radius is 5

the furthest distance of y radius is 4

but the centre is (0,-2)

then for the point of intersection...

just use x = 0 and y= 0

and just find it

like

0 = 5cost

and 1/2 = sin t

use your calculator to find it

it isnt something hard.

thats all i can say.

i hope i aint wrong..cause ive tried using this way in almost all of the past year questions and got most of it right

i agree that the parametric gives an elipse shape

but i dont think its in terms of pi

x = 5 cos t, y = -2 + 4 sin t

the furthest distance of x radius is 5

the furthest distance of y radius is 4

but the centre is (0,-2)

then for the point of intersection...

just use x = 0 and y= 0

and just find it

like

0 = 5cost

and 1/2 = sin t

use your calculator to find it

it isnt something hard.

thats all i can say.

i hope i aint wrong..cause ive tried using this way in almost all of the past year questions and got most of it right

MalaysianDude

erm sorry if im wrong or something but

i agree that the parametric gives an elipse shape

but i dont think its in terms of pi

x = 5 cos t, y = -2 + 4 sin t

the furthest distance of x radius is 5

the furthest distance of y radius is 4

but the centre is (0,-2)

thats all i can say.

i hope i aint wrong..cause ive tried using this way in almost all of the past year questions and got most of it right

i agree that the parametric gives an elipse shape

but i dont think its in terms of pi

x = 5 cos t, y = -2 + 4 sin t

the furthest distance of x radius is 5

the furthest distance of y radius is 4

but the centre is (0,-2)

thats all i can say.

i hope i aint wrong..cause ive tried using this way in almost all of the past year questions and got most of it right

It is in terms of pi, but you are right the centre is (0,-2). Because the next question asked for a sketch of the ellipse. And the centre is (0, -2) and the radiuses are 5 and 4.

then just let x as 0

find the value of t and put it into the other equation to get the y value

then repeat it but this time use y as 0

find the value of t and put it into the other equation to get the y value

then repeat it but this time use y as 0

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