# P5 Ellipses

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#1
Its q.16 p.88 edexcel p5

show that S(root3,0) is a focus of the ellipse eq/n 3x^2 + 4y^2 = 36. O is origin, P is a pt. on ellipse, a line is draw from O perp. to the tangent to the ellipse at P and this line meets the line SP (produced if neccessary) at the pt. Q. Show that the locus of Q is a circle.

can do the proof bit but not the locus job, help wud be good than q
0
15 years ago
#2
(Original post by Gregball_87)
Its q.16 p.88 edexcel p5

show that S(root3,0) is a focus of the ellipse eq/n 3x^2 + 4y^2 = 36. O is origin, P is a pt. on ellipse, a line is draw from O perp. to the tangent to the ellipse at P and this line meets the line SP (produced if neccessary) at the pt. Q. Show that the locus of Q is a circle.

can do the proof bit but not the locus job, help wud be good than q
ill have a go. sorry if it contans errors too messy doing on machine.
let P=(x1,y1)
from eqn of ellipse 3x+4ydy/dx=0
dy/dx=-3x1/4y1 at P
line perp to tangent at P has grad 4y1/3x1
since this line passes through (0,0) eqn of line OT,say, is y=4y1x/3x1
eqn of SP is given by
y=y1(x-rt(3))/(x1-rt(3))
the two lines meet when
4y1x/3x1=y1(x-rt(3))/(x1-rt(3))
(x1-4rt(3))x=-3rt3x1
ie x=-3rt(3)x1/(x1-4rt(3)) ....................(1)
this gives y=-4rt(3)y1/(x1-4rt(3))..........(2)
(1) gives x1x-4rt(3)x+3rt(3)x1=0
x1(x+3rt(3))=4rt(3)x
x1=4rt(3)x/(x+3rt(3)).....................( 3)
using (3) gives
x1-4rt(3)=4rt(3)x/(x+3rt(3))-4rt(3)=(4rt(3)x-4rt(3)x-36)/(x+3rt(3))
=-36/(x+3rt(3))...................... .(4)
putting (4) into (2) gives
y=rt(3)y1(x+3rt(3))/9
so 9y/(rt(3)(x+3rt(3))=y1............. ..........(5)
since x1 y1 lie on ellipse they satisfy eqn of ellipse this gives
3.16.3x^2/(x+3rt(3))^2+81.4y^2/3(x+3rt(3))^2=36
144x^2+108y^2=36(x+3rt(3))^2
=36x^2+6.36rt(3)x+27.36
108x^2-216rt(3)x+108y^2=972
x^2-2rt(3)+y^2=9
(x-rt(3))^2+y^2=12 well theres hope, that is eqn of circle
0
#3
jesus wept, no wonder i struggled

thanks alot
0
15 years ago
#4
This question is about ellipses. I'd be grateful if someone could shed some light on where my solution went wrong:

An ellipse has focus S (rt5, 0) and equation (x^2)/9 + (y^2)/4 = 1
The variable point T(3cost, 2sint) is joined to S. The line ST is produced to P so ST/SP = (1/3). Find the locus of P as t varies.

Firstly i rearranged the ratio expression to give 3ST=SP.
First i tried to find a parametric expression for ST.
ST: x=3cost-rt5. y=2sint (as ST = T - S?)
SP: x=9cost-3rt5. y=6sint
cost = (x+3rt5)/9. sint=y/6
(x+3rt5)^2 /81 + y^2/36 = 1
4(x+3rt5)^2 + 9y^2 = 324.
However, the book give the same answer but the coefficient of rt5 is 2.
I'd greatly appreciate any help.
0
15 years ago
#5
(Original post by Gaz031)
This question is about ellipses. I'd be grateful if someone could shed some light on where my solution went wrong:

An ellipse has focus S (rt5, 0) and equation (x^2)/9 + (y^2)/4 = 1
The variable point T(3cost, 2sint) is joined to S. The line ST is produced to P so ST/SP = (1/3). Find the locus of P as t varies.

Firstly i rearranged the ratio expression to give 3ST=SP.
First i tried to find a parametric expression for ST.
ST: x=3cost-rt5. y=2sint (as ST = T - S?)
SP: x=9cost-3rt5. y=6sint
cost = (x+3rt5)/9. sint=y/6
(x+3rt5)^2 /81 + y^2/36 = 1
4(x+3rt5)^2 + 9y^2 = 324.
However, the book give the same answer but the coefficient of rt5 is 2.
I'd greatly appreciate any help.
x cord of t is 3cost so x-cord of p is 9cost-2rt(5)
eg if S was at (2,0) and P at (5,4) then SP=rt(9+16)=5 so we would need ST to have length 15 so T would have coords (11,12) so ST has length rt(81+144)=15.
ie x-cord of T is 3xcord of P- 2.2
y cord of T is 6sint
x=9cost-2rt(5)
gives 2(x+2rt(5))=18cost
3y=18sint
so 4(x+2rt(5))^2+9y^2=324.
hope this ok in a rush in a lecture.
0
15 years ago
#6
Thanks. I understand now. It helps me to think of it more from the vector side of things. Ie: First find the change from S to T, then from S to T, finally add the position vector of S to get the final coordinates.
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