ind 2008 q4 b) ii) (1 Viewer)

[duy.le]

ok this question has got to be one of the weirdest conics question, i normally always get them but this is wtf. i know i must have a wrong method cause the question is only worth 3 marks...

P(a cos @, b sin @) is a point on the ellipse [insert generic equation]. PQ is a diameter of the ellipse. The tangent to the ellipse at P meets the vertical through Q at R.

i) prove the equation for the tangent at p. (cbb writing it, u guys should know it off by heart by now anyways so moving on...) [parametric form]

ii) Show that the ratio of the area of triangle PQR to the area of the ellipse is 2: pi |tan @|

help. if u have the actual answer could u post them up (the whole paper if u already have it in digital form)

Thanks

 

[friction]

hmm i did this and had the answers and got it right but i think i threw it out sorry.

I just remember putting it all together and there it was.

 

[hon1hon2hon3]

diameter of a ellipse is any line that goes through the foci ? how can the tangent go through the diameter . . dont get it = ="

 

[independantz]

diameter of a ellipse is any line that goes through the foci ? how can the tangent go through the diameter . . dont get it = ="

Nah the diameter, is an chord that passes through the centre of the ellipse.

 

[Affinity]

Q has coordinates (-acos@,-bsin@)

the tangent line at P is

(x - acos@)(-b cos@)/(a sin@) = (y - bsin@)

simplifies to

(a sin@)y + (b cos@)x = ab

now R = (-acos@, b(1 + cos^2 @)/sin@)

So QR = [b(1 + cos^2 @)/sin@ ]+ bsin@ = 2b/sin @

using that as base, the correspnding height would be

2acos@

so the area will be 2ab/tan@

the area of the ellipse is pi a b

so the ratio is 2/[pi|tan(@)|]

you should be able to fill in the details

 

[duy.le]

Q has coordinates (-acos@,-bsin@)

the tangent line at P is

(x - acos@)(-b cos@)/(a sin@) = (y - bsin@)

simplifies to

(a sin@)y + (b cos@)x = ab

now R = (-acos@, b(1 + cos^2 @)/sin@)

So QR = [b(1 + cos^2 @)/sin@ ]+ bsin@ = 2b/sin @

using that as base, the correspnding height would be

2acos@

so the area will be 2ab/tan@

the area of the ellipse is pi a b

so the ratio is 2/[pi|tan(@)|]

you should be able to fill in the details

OMG i cant believe i didnt see this, wow thank you sooo much.

 

[samwell]

Do we have to show how we got the area of the ellipse coz its like onli 3 marks. I noe the area is pi (ab) but u wouldnt wanna find the area using integration for only a 3 marker.
Does anybody have another method to find area of an ellipse in a quicker and shorter way???

 

[duy.le]

i think that is the simplest way, i would say ur allowed to assume it, proving it is long (relatively)

 

[Zeber]

Do we have to show how we got the area of the ellipse coz its like onli 3 marks. I noe the area is pi (ab) but u wouldnt wanna find the area using integration for only a 3 marker.
Does anybody have another method to find area of an ellipse in a quicker and shorter way???

I'd say you can assume it, but if you wanna prove it, it's only like 4-5 lines of working considering the integral to be a quarter of a circle (it is in the form a^2-x^2 if i recall correctly)