maths help

I had a look and I didn't do it by substituting k in right away. What does OA tell you about the relationship between the x and y coordinates of A? Given the circle equation and A lies on it, there's only one point that satisfies both (given k>0). But I suppose the substitution first would get you there as well.

Reply 2

OP

Original post by *****deadness

I had a look and I didn't do it by substituting k in right away. What does OA tell you about the relationship between the x and y coordinates of A? Given the circle equation and A lies on it, there's only one point that satisfies both (given k>0). But I suppose the substitution first would get you there as well.

in the OA vector, j moves the y coordinate and i moves the x coordinate?

Yeah

OP

Original post by *****deadness

Yeah

if you don't substitute it first, what would you do?

Reply 5

Well OA=2ki+kj, so without giving the method away, this directly tells me the ratio between the x and y coordinate of A. Then rewriting y in terms of x at A (or x in terms of y if you like), substitute into the circle equation to find for which point on the circle it has that ratio

Reply 6

OP

Original post by *****deadness

Well OA=2ki+kj, so without giving the method away, this directly tells me the ratio between the x and y coordinate of A. Then rewriting y in terms of x at A (or x in terms of y if you like), substitute into the circle equation to find for which point on the circle it has that ratio

well I think I have done this completely wrong but I've done so far
let i = x and j = y
2kx+ky=0
ky=-2kx
y=-2kx/k
substitute y into the circle equation
x^2-2kx/k=9
I'm I going in the wrong path?

Reply 7

Slightly...since we're going from the origin O, we've moved 2ki+kj to A, which means moving 2k in the positive x direction and k in the positive y direction. And since we started from O, A is (2k,k). Now this can be anywhere since k is a constant, but there is only 1 of these points that also lies on the circle

Reply 8

OP

Original post by *****deadness

Slightly...since we're going from the origin O, we've moved 2ki+kj to A, which means moving 2k in the positive x direction and k in the positive y direction. And since we started from O, A is (2k,k). Now this can be anywhere since k is a constant, but there is only 1 of these points that also lies on the circle

since k=p√5, what do I do to find what p is? I understand what the question is telling me to do but I'm unsure on how to solve it?

Reply 9

OP

Original post by *****deadness

Slightly...since we're going from the origin O, we've moved 2ki+kj to A, which means moving 2k in the positive x direction and k in the positive y direction. And since we started from O, A is (2k,k). Now this can be anywhere since k is a constant, but there is only 1 of these points that also lies on the circle

oh wait, considering A is (2k,k) do I substitute x=2k and y=k into the circle equation?

Reply 10

Yeah and you'll get what k is which tells you what p is

Reply 11

OP