conics question (1 Viewer)

[siavash_s_s]

hey
due to my time wasting in assorted activities (setting up servers with j0n ) i am now lost to a certain degree, so full steam ahead

say we have 2 points
P (aCos , bSin ) and Q (aCos -, bSin -)
and the ellipse
x^2/a^2 + y^2/b^2 = 1

we want to see if PQ passes through (0,0)

what im thinking is, instead of finding PQ and substituting it in, its quicker to use the thing that gets the middle of a line.. cant remember the name. that would give us the centre of the line at (0,0), but would you get the mark for assuming it is then proving it with that?

i just re read that and didnt understand... oh well

same thing for a similar question for hyperbola

 

[siavash_s_s]

hey does anyone here have a link or a document or pdf of cambridge worked solutions, a friend told me some people here do

 

[KeypadSDM]

Originally posted by siavash_s_s
hey does anyone here have a link or a document or pdf of cambridge worked solutions, a friend told me some people here do

Your friend is lying. (Unless I'm horribly mistaken)

 

[KeypadSDM]

Originally posted by siavash_s_s
hey
due to my time wasting in assorted activities (setting up servers with j0n ) i am now lost to a certain degree, so full steam ahead

say we have 2 points
P (aCos , bSin ) and Q (aCos -, bSin -)
and the ellipse
x^2/a^2 + y^2/b^2 = 1

we want to see if PQ passes through (0,0)

what im thinking is, instead of finding PQ and substituting it in, its quicker to use the thing that gets the middle of a line.. cant remember the name. that would give us the centre of the line at (0,0), but would you get the mark for assuming it is then proving it with that?

i just re read that and didnt understand... oh well

same thing for a similar question for hyperbola

If you show the midpoint is (0,0) for all cases, then it has to pass through the point (0,0). A perfectly legit proof, if it works:

P (aCos , bSin ) and Q (aCos -, bSin -)

But you forget, these two points lie in adjacant quadrants.

 

[George W. Bush]

Clearly PQ won't pass through the origin.

 

[KeypadSDM]

Originally posted by George W. Bush
Clearly PQ won't pass through the origin.

Thanks for stating the obvious. Maybe you can do it in your next address to the nation? We all know the American population needs it

 

[xiao1985]

Originally posted by George W. Bush
Clearly PQ won't pass through the origin.

no, in most cases they won't...

 

[BlackJack]

bah.
State that as the y-values of P and Q are of equal magnitude and opposite in sign, the midpoint will always have y-value 0.
X values are identical across the board, hence midpoint will have value X. And so PQ is parallel to y-axis, by observation.
Hence, PQ passes through (0,0) if and only if X is 0...


unless, o.c., that theta stands for different values.

 

[siavash_s_s]

i made possibly the dumbest mistake in the world
i wrote:
Q (aCos -, bSin -)

inplace of Q(aCos ( + pi), bSin ( + pi) i wrote the above instead of Q (-aCos , -bSin )

i hope what i said just then is right.. i hate lookn dumb