C4 - Parametric equation

Need help with 9iii.

Crosses x-axis when y=0 and crosses y-axis when x=0

Crosses x-axis 4ysinp=12

y = 12/ 4sinp --> 3sinp

Crosses y-axis 3xcosp=12
x =12/3cosp =4cosp

Apparently it is 3/sinp and 4/cosp

Can someone explain why?

(edited 8 years ago)

C4 - Parametric equation.jpg49.7KB

Reply 1

davros

16

Original post by Super199

Need help with 9iii.

Crosses x-axis when y=0 and crosses y-axis when x=0

Crosses x-axis 4ysinp=12

y = 12/ 4sinp --> 3sinp

Crosses y-axis 3xcosp=12
x =12/3cosp =4cosp

Apparently it is 3/sinp and 4/cosp

Can someone explain why?

I think you need to re-learn division :smile:

12/(4sinp) = 3/sin p

and

12/(3cosp) = 4/cos p

:smile:

Reply 2

Super199

OP

20

Original post by davros

I think you need to re-learn division :smile:

12/(4sinp) = 3/sin p

and

12/(3cosp) = 4/cos p

:smile:

Yeah I have just realised :tongue:

Reply 3

joe1545

14

Original post by Super199

Need help with 9iii.

Crosses x-axis when y=0 and crosses y-axis when x=0

Crosses x-axis 4ysinp=12

y = 12/ 4sinp --> 3sinp

Crosses y-axis 3xcosp=12
x =12/3cosp =4cosp

Apparently it is 3/sinp and 4/cosp

Can someone explain why?

So when when the tangent crosses the x axis, y=0, which will give us 3xCosP=12

rearrange for x and we get x = 12/3(SecP) = 4SecP

Do the same to find the y intercept and we should get y = 3CosecP

Now if you draw oout a right angled triangel, you'll see that the base of the triangle (ie O to the x intercept) = 4SecP/

And the height is 3CosecP.

So the area will be 1/2 * 3/SinP * 4/CosP = 12/2SinPCosP

Now using the double angle formula for Sin2P it is then apparent that the area = 12/Sin2P.

In answer to your question, when you rearrange to find x and y, you're dividing 12, by 3CosP = 12/3CosP =4/CosP

(edited 8 years ago)

Reply 4

Super199

OP

20

Original post by joe1545

So when when the tangent crosses the x axis, y=0, which will give us 3xCosP=12

rearrange for x and we get x = 12/3(SecP) = 4SecP

Do the same to find the y intercept and we should get y = 3CosecP

Now if you draw oout a right angled triangel, you'll see that the base of the triangle (ie O to the x intercept) = 4SecP/

And the height is 3CosecP.

So the area will be 1/2 * 3/SinP * 4/CosP = 12/2SinPCosP

Now using the double angle formula for Sin2P it is then apparent that the area = 12/Sin2P.

In answer to your question, when you rearrange to find x and y, you're dividing 12, by 3CosP = 12/3CosP =4/CosP

Yeah I have completed this question but thanks :smile:

Do you mind helping me with another question actually.

6ii. Parallel to y-axis the bottom is =0

2xy-3 = 0
xy =3/2 Care to explain how that shows no tangents are parallel to y-axis. Is it because xy cannot ever be equal to 3/2. Not entirely sure... C4 ....jpg

Reply 5

lizard54142

6

Original post by Super199

Yeah I have completed this question but thanks :smile:

Do you mind helping me with another question actually.

6ii. Parallel to y-axis the bottom is =0

2xy-3 = 0
xy =3/2 Care to explain how that shows no tangents are parallel to y-axis. Is it because xy cannot ever be equal to 3/2. Not entirely sure... C4 ....jpg

xy=\frac{3}{2} \Rightarrow x = \frac{3}{2y}

Use the equation of the curve...

Reply 6

joe1545

14

Original post by Super199

Yeah I have completed this question but thanks :smile:

Do you mind helping me with another question actually.

6ii. Parallel to y-axis the bottom is =0

2xy-3 = 0
xy =3/2 Care to explain how that shows no tangents are parallel to y-axis. Is it because xy cannot ever be equal to 3/2. Not entirely sure... C4 ....jpg

Yeah, i'll work through it quickly now, although bare in mind, that when a tangent to a curve is parallel to the x axis, we say the dy/dx = 0. Hence in this case the numerator would be = 0

Although when a tangent is parralel to the y axis, we say y tends towards infinity for all values of x. T
This means that the denominator would have to be = 0.

Hence 2xy-3 = 0

Reply 7

Super199

OP

20

Original post by lizard54142

xy=\frac{3}{2} \Rightarrow x = \frac{3}{2y}

Use the equation of the curve...

What am I supposed to see sorry? When I sub it in?

Reply 8

joe1545

14

Original post by Super199

Yeah I have completed this question but thanks :smile:

Do you mind helping me with another question actually.

6ii. Parallel to y-axis the bottom is =0

2xy-3 = 0
xy =3/2 Care to explain how that shows no tangents are parallel to y-axis. Is it because xy cannot ever be equal to 3/2. Not entirely sure... C4 ....jpg

Im about to go and do a past paper now, but as said, rearrange for y, and sub back into the curve. You should then be able to get a quadratic equation which you can then see if there are any real roots too. If there are no real roots then you know there arent any tangents parrallel to the y axis.