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C4- Integration question with parametric equations

Edit: Ignore this post. made stupid mistake in calculations

C4 Heinemann Edexcel Mixed Exercise 6L question 15.

The curve shown has parametric equations
x=5cosθ,y=4sinθ x = 5 \cos \theta, y = 4 \sin \theta

(The curve is basically an ellipse centred on the origin)

a) Find the gradient of the curve at the point P at which pheta = π/4
b) Find an equation of the tangent to the curve at the point P
c) Find the coordinates of the point R where this tangent meets the x-axis.

Got the first three- easy but it's question d. i'm stuck on.

The shaded region is bounded by the tangent PR, the curve and the x-axis.

d) Find the area of the shaded region, leaving your answer in terms of π.

I know that the tangent meets the x-axis outside the ellipse, to the right of it.
So what I tried was to find the area of the right-angled triangle under PR and subtract from that the definate integral between pheta = 0 and pheta = π/4...

This is the part I'm a bit dubious about because I'm not sure what area the integral would give me. Am I right to use this method? Would it give me the area of the ellipse above the x-axis or below it as well.

Thanks in advance. :smile:

Reply 1

Integrate the equation of the curve finding the area inside the ellipse, and subtract this from the area of OPR (the triangle).

Reply 2

Original post by biatch
Edit: Ignore this post. made stupid mistake in calculations

C4 Heinemann Edexcel Mixed Exercise 6L question 15.

The curve shown has parametric equations
x=5cosθ,y=4sinθ x = 5 \cos \theta, y = 4 \sin \theta

(The curve is basically an ellipse centred on the origin)

a) Find the gradient of the curve at the point P at which pheta = π/4
b) Find an equation of the tangent to the curve at the point P
c) Find the coordinates of the point R where this tangent meets the x-axis.

Got the first three- easy but it's question d. i'm stuck on.

The shaded region is bounded by the tangent PR, the curve and the x-axis.

d) Find the area of the shaded region, leaving your answer in terms of π.

I know that the tangent meets the x-axis outside the ellipse, to the right of it.
So what I tried was to find the area of the right-angled triangle under PR and subtract from that the definate integral between pheta = 0 and pheta = π/4...

This is the part I'm a bit dubious about because I'm not sure what area the integral would give me. Am I right to use this method? Would it give me the area of the ellipse above the x-axis or below it as well.

Thanks in advance. :smile:

I know I’m replying to a message that was years ago but couldn’t help notice no solution was found and I was trying to do this problem so hopefully my method I attached will help someone looking for the way to go about this question.

Reply 3

Original post by JJW1643
I know I’m replying to a message that was years ago but couldn’t help notice no solution was found and I was trying to do this problem so hopefully my method I attached will help someone looking for the way to go about this question.

How did you find the equation of the eclipse to integrate?

Reply 4

Original post by JJW1643
I know I’m replying to a message that was years ago but couldn’t help notice no solution was found and I was trying to do this problem so hopefully my method I attached will help someone looking for the way to go about this question.

How did you find the equation of the eclipse to integrate?

Reply 5

Original post by JJW1643
I know I’m replying to a message that was years ago but couldn’t help notice no solution was found and I was trying to do this problem so hopefully my method I attached will help someone looking for the way to go about this question.

Where did the (4sin q)(-5sin q) come from?

*let theta equal q

Reply 6

Where did the (4sin q)(-5sin q) come from?

*let theta equal

Reply will be useless to you now, but for anyone else:
The area = the integral of y with respect to x (dx). So use y=,,,,, and then differentiate x to get dx and time together and find the integral of this for the area.

Reply 7

Original post by JJW1643
I know I’m replying to a message that was years ago but couldn’t help notice no solution was found and I was trying to do this problem so hopefully my method I attached will help someone looking for the way to go about this question.


your a life saver, thank you