Chord of an ellipse (1 Viewer)

[js992]

The question is



Letting (θ+ pi) = ɸ
I found the gradient of the chord, then the equation of it, cleaned it up but ended up getting


giving



but as ɸ = (θ+ pi)
the RHS is then

whereas i think it should be

I don't know what i did wrong or what i'm meant to do next

Edit : This 4U Cambridge 3.2 Q3 btw

 

[life92]

Okay I'm not quite sure what you did, but what I got was MUCH simpler.

Firstly, for the gradient of PQ
m = {b [ sin(theta+pi)-sintheta ] } / {a [cos(theta+pi) - costheta] }
and then expanding out the sin(theta+pi) and cos(theta+pi)
it simplifies to
m = b sin theta / a cos theta

Then subbing into point-gradient formula.

y - bsin theta = (b sin theta / a cos theta) (x - acos theta)
which simplifies to
aycos theta = bxsin theta
which clearly passes through (0,0)

If you don't know how I simplified, just ask and I'll type it up here.

PS. sorry ! I still don't know how to use Latex

 

[js992]

Oh that is much easier -_-

I dunno, i had a look at the worked cambridge solutions
They used the identity that


and you get some huge mess which sort of cleans itself up but i get stuck halfway.

The solutions also skip 4000000 steps so it confused me

 

[life92]

I see.
I worked it out using that sums to product formula, and it comes out the same.
The amount of working is around the same, but knowing to use sums to product in an exam... I would only advise it if you have used it alot and are confident with it and know trig ratios well.

For example, in this question, if you use the sums to product formula.

You get cos [ (2theta+pi)/2 ]
= cos (theta + pi/2)
= cos [pi - (pi/2 -theta)]
= - cos (pi/2 - theta)
= - sin theta

So yes, just remember there are often several methods to solving one question, especially in conics questions.

 

[cutemouse]

Those formulas used to be in the old 2 unit course, so I guess I wouldn't hurt memorising them.

 

[js992]

Ahhh i see, still your way is much simpler haha

Thanks alot

 

[life92]

With the product to sums and sums to products formulae, yeah it wouldn't hurt to memorise them especially for applications in Conics and possibly Complex and Polys.

But since they were part of the old course, its unlikely you'd be expected to need to be able to use them.

Being able to prove/derive them is more important, because questions like,
"i)Prove this...ii)Hence solve/integrate/show..." are more likely to come up.

PS. For questions like these where there are more than one method to solving it, try solving it the various ways, and see which method suits you best. By doing this you can find which method/approach will be best for you in an exam.

 

[cutemouse]

it wouldn't hurt to memorise them especially for applications in Conics and possibly Complex and Polys.

and integration.

 

[cyl123]

um... just change Q[acos(pi+theta), bsin(pi+theta)] to Q[-acos(theta), -bsin(theta)] with basic 2U trig and its pretty obvious from there. No need for sum or differences of trig formula.

 

[brittricho]

Have you noticed how confusing the "worked solutions" are in that textbook?? They do things that aren't 'obvious', and don't explain what they are doing!! It's really frustrating when you have no idea how to approach a question!!