Conics Problems (1 Viewer)

[KeypadSDM]

Originally posted by wildtiger
hi all - i have a problem with a question in cambridge's CONICS chapter on page 92, would anyone be able to help me out? Thanks!

I'm stuck on question 7b)

7. P(a cos A, b sin B) and Q (a cos A, b sin B)
lie on the ellipse x^2/a^2 + y^2/b^2 =1.

a) If PQ subtends a right angle at (0,0), show that tan A tan B = - a^2/b^2

b) If PQ subjects a right angle at (a, 0) show that tan (A/2) tan (B/2) = - b^2/a^2

 

[ND]

I'm guessing these q's are for the '04ers?

 

[KeypadSDM]

Why not let them have a go. I think they already know we're good.

 

[hatty]

u guys have trouble solving this?
LOL

 

[freaking_out]

btw, fully worked solutions to the cambridge textbook is available at drbuchanan's website.

 

[:: ck ::]

not many schools have finished complex no's or even started 4unit ...

maybe one of u 03'ers would be nice enough to help out =)

 

[Giant Lobster]

i can do that question. < /brag>

 

[ND]

Originally posted by hatty
u guys have trouble solving this?
LOL

We were leaving it for the '04ers.

not many schools have finished complex no's or even started 4unit ...

maybe one of u 03'ers would be nice enough to help out =)

You don't really need to have done conics for this q. All you need to know are the polar coords (which are given), and that 'a' is the pt of intersection of the ellipse and the positive x-axis.

Though to do this q, we need to know what A and B represent.

 

[J0n]

Originally posted by ND
Though to do this q, we need to know what A and B represent.

Looking at the book, i would say that A and B represent X and Y respectively.

 

[ND]

Ok then, here are some tips:

Y=X+pi/2 in the 1st one
also, tanY=-tan(pi-Y)

edit: actually, it looks as though teh 1st one comes out to be -b^2/a^2. =/

 

[wildtiger]

Sorry.

7. P(a cos X, b sin X) and Q (a cos Y, b sin Y)

should be

7. P(a cos A, b sin B) and Q (a cos A, b sin B)

i dont know why i switched from As and Bs ==> Xs and Ys... sorry to those who wasted their time >_< trying to solve this using As and Bs

 

[Rorix]

Originally posted by ND
Ok then, here are some tips:

Y=X+pi/2 in the 1st one
also, tanY=-tan(pi-Y)

edit: actually, it looks as though teh 1st one comes out to be -b^2/a^2. =/

Yes, this is what I got too.....

(note: I have never done conics before so my methods may be slow/wrong. Go easy)

Assuming P(acosA, bsinA) and Q (acosB, bsinB) because wildtiger's new versions are identical

Gradient OP= b/a tanA
Gradient OQ = b/a tanB

OP and OQ are perpendicular so b/a tanA b/a tanB = -1
b^2/a^2 tanAtanB = -1
Oh, there we go
tanAtanB = -a^2/b^2

Excuse my newbiness, but what is the subjected angle? A google search didn't offer anything. From what ND said I think I have an idea so I'll give it a shot.

Gradient AP = b/a (sinA/1-cosA)
Gradient AQ = b/a (sinB/1-cosB)

Using half angle simplifies to
Grad AP = b/a (-cot A/2)
Grad AQ = b/a (-cot B/2)

Once again lines are perpendicular so b^2/a^2 cotA/2cotB/2 = -1
cotA/2cotB/2 = -a^2/b^2
tanA/2tanB/2 = -b^2/a^2

Mise.