Guys i seriously need help with conics (1 Viewer)

[currysauce]

I am not understanding this.... well i get it... but i can do about half the question b4 not being able to finish it...

specifcally, if anyone has the cambridge book, excercise 3.2, qu 9b and ex 3.3, qu 4,5,6,7,

if anyone can help with any of these i would be greatly thankful...

i'm considering dorpping 4u... i mean its really hard

anyway thanks

 

[breaking]

frustrum of a cone!!!!!!!!!!!!!!!!!!!!

 

[McLake]

If you can post the question ...

 

[currysauce]

sure.

P(2root3, 3root3) is one extremeity of a focal chord on the hyperbola x^2/3 - y^2/9 = 1 . Find the coordinates of the other extrmiety Q.

2.

P(acos(theta), bsin(theta)) lies on the ellipse (normal ellipse. eqn.) The normal at P cuts the x-axis at X and the y-axis at Y. Show that PX/PY = b^2/a^2.

Suprisingly, this questoin^^ i can only get halfway before getting stuck with algebra/

3.

P(asec(theta), btan(theta)) lies on the hyperbola (normal hy. eqn) The tangent at P cuts the x-axis at X and y and Y. Show that PX/PY = sin^2(theta) and deduce that if P is an extremity of the latus rectum, then PX/PY = e^-1/e^2.


i'll post more later


thanks mclake if u can spare the trouble. or anyone else

 

[withoutaface]

I seriously wouldn't worry if you can't do those conics questions in cambridge.

 

[McLake]

I am with withoutaface on this one. Have a go at easier conics questions first. If you already have, and you were fine with those, then leave these until later.

As I posted somewhere eles, there are basically three standard conics questions:

- Find the tangent/normal
- Find a locus
- Prove/Use a conics rule/property

The questions above are Q8 candidates, try tackling them later in the year. If you want help anyway, then let me know and I will have a go.

 

[Slidey]

Don't drop because it is hard now. Osmosis, among other things, will ensure it becomes more thoroughly cemented in your head by the time the HSC exams come around.

 

[currysauce]

why shouldn't i worry, if i dont get something done i'll be worrying about it till it is

hence, i'll be worrying all my holidays

 

[Ogden_Nash]

1. This question is similar to the example in the book a few pages earlier.
P(2√3,3√3) = P(√3sec(π/3),√3tan(π/3))
∴ P has parameter π/3. Let Q have parameter φ.
Now, from part a) of the question, tan(θ/2)tan(φ/2) = (1-e)/(1+e) or (1+e)/(1-e)
∴ tan(π/6)tan(φ/2) = (1-2)/(1+2) or (1+2)/(1-2)
ie tan(φ/2) = - 1/√3 or -3√3
Now, using the t results for tan(φ/2):
sec( φ ) = (1-1/3)/(1+1/3) = 2 or (1+27)/(1-27) = -14/13 and similarly
tan( φ ) = -√3 or 3√3/13
∴ Q(√3secφ,3tanφ ) => Q(2√3,-3√3) OR Q(-14√3/3,9√3/13). Hope that helps, try and understand every step of the solution.