# maths help

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

Given that k = p√5 and that p > 0, state the value of p as a fraction in its lowest terms.

The point A lies on the circle with equation x^2+y^2=9 and vector OA = 2ki + kj

so far I substituted k into the vector so its OA=2p√5i+p√5j but then I don't know what to do next.

The point A lies on the circle with equation x^2+y^2=9 and vector OA = 2ki + kj

so far I substituted k into the vector so its OA=2p√5i+p√5j but then I don't know what to do next.

0

reply

Report

#2

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.

0

reply

(Original post by

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.

*******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.

0

reply

(Original post by

Yeah

*******deadness**)Yeah

0

reply

Report

#6

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

0

reply

(Original post by

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

*******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

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?

0

reply

Report

#8

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

0

reply

(Original post by

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

*******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

0

reply

*******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

0

reply

(Original post by

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

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

2k^2+k^2-9=0

k=-1+√73/4 or -1-√73/4

considering p>0, k=-1+√73/4

divide k by √5 to find p and i get p=√365-√5/20

it's quite a wacky number so im not sure if im correct?

0

reply

Report

#14

And even with 3k^2-9=0 what went so wrong that you got that? Even with this one it's k=-root 3 or +root 3 right

0

reply

(Original post by

And even with 3k^2-9=0 what went so wrong that you got that? Even with this one it's k=-root 3 or +root 3 right

*******deadness**)And even with 3k^2-9=0 what went so wrong that you got that? Even with this one it's k=-root 3 or +root 3 right

0

reply

(Original post by

Yeah it's not what I got, which gives me p as an actual fraction

*******deadness**)Yeah it's not what I got, which gives me p as an actual fraction

0

reply

Report

#19

I said that because I was confused how you got the ugly surds even with 3k^2-9=0, but it should really be (2k)^2+(k)^2-9=0

0

reply

(Original post by

I said that because I was confused how you got the ugly surds even with 3k^2-9=0, but it should really be (2k)^2+(k)^2-9=0

*******deadness**)I said that because I was confused how you got the ugly surds even with 3k^2-9=0, but it should really be (2k)^2+(k)^2-9=0

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top