# maths help

Watch
Announcements
#1
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.
0
1 month ago
#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
#3
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?
0
1 month ago
#4
Yeah
0
#5
Yeah
if you don't substitute it first, what would you do?
0
1 month ago
#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
#7
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?
0
1 month ago
#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
#9
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?
0
#10
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?
0
1 month ago
#11
Yeah and you'll get what k is which tells you what p is
0
#12
Yeah and you'll get what k is which tells you what p is
so I did
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
1 month ago
#13
Yeah it's (2k)^2
0
1 month ago
#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
1 month ago
#15
Yeah it's not what I got, which gives me p as an actual fraction
0
#16
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
I accidentally did 2k^2+k-9 instead of 3k^2-9 oops
0
#17
Yeah it's not what I got, which gives me p as an actual fraction
I ended up with p=√15/5?
0
1 month ago
#18
NO it's actually (2k)^2
0
1 month ago
#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
#20
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
finally got it now😭😭, thank you so much for your help
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Should there be a new university admissions system that ditches predicted grades?

No, I think predicted grades should still be used to make offers (716)
33.9%
Yes, I like the idea of applying to uni after I received my grades (PQA) (904)
42.8%
Yes, I like the idea of receiving offers only after I receive my grades (PQO) (399)
18.89%
I think there is a better option than the ones suggested (let us know in the thread!) (93)
4.4%