• Want to help us with this year's BoS Trials?
    Let us know before 30 June. See this thread for details
  • Looking for HSC notes and resources?
    Check out our Notes & Resources page

Conics help (1 Viewer)

clintmyster

Prophet 9 FTW
Joined
Nov 12, 2007
Messages
1,067
Gender
Male
HSC
2009
Uni Grad
2015
For the a part, im a bit iffy about my way in doing part (IV) yet but il keep trying. cbb to b) but im sure if you manipulate around a bit you'l get there.

im sure you can show that formula
ii) it seems a gift IMO. sub in (ae,0) and you'l get it
iii) i used y2 - y1 / x2 - x1 and you notice how the denom is 0 therefore the gradient is infinite. Therefore the angle must be 90 degrees
iv) Use similar triangles ROT and PST
therefore RT/RO = PT/PS
therefore RT ( 1 - e^2) = PT
PR = RT - PT
= RT - (RT [ 1 - e^2 ] )
= e^2 . RT


seems i could be bothered to do the part b)

as i said state eqns of tangent/normal.
to find T and N, replace y in both with x
so the coordinates of T are (2cp/(1+p^2), 2cp/(1+p^2)) and N are (c(p^2 +1)/p , c(p^2 +1)/p)
now to find OT and ON use pythagoras taking the x and y coordinates of T and N
multiply OT and ON and heaps of stuff cancels and you left with 4c^2
 
Last edited:

study-freak

Bored of
Joined
Feb 8, 2008
Messages
1,133
Gender
Male
HSC
2009
For the a part, im a bit iffy about my way in doing part (IV) yet but il keep trying. cbb to b) but im sure if you manipulate around a bit you'l get there.

im sure you can show that formula
ii) it seems a gift IMO. sub in (ae,0) and you'l get it
iii) i used y2 - y1 / x2 - x1 and you notice how the denom is 0 therefore the gradient is infinite. Therefore the angle must be 90 degrees
iv) Use similar triangles ROT and PST
therefore RT/RO = PT/PS
therefore RT ( 1 - e^2) = PT
PR = RT - PT
= RT - (RT [ 1 - e^2 ] )
= e^2 . RT


seems i could be bothered to do the part b)

as i said state eqns of tangent/normal.
to find T and N, replace y in both with x
so the coordinates of T are (2cp/(1+p^2), 2cp/(1+p^2)) and N are (c(p^2 +1)/p , c(p^2 +1)/p)
now to find OT and ON use pythagoras taking the x and y coordinates of T and N
multiply OT and ON and heaps of stuff cancels and you left with 4c^2
sub T(a/e,0)
 

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,178
Gender
Male
HSC
2006
That's the bit I don't get though. How do I know that the Tangent passes through (a/e, 0)?
Read the question. It says that point T intersects with the directrix and because its an x-intercept, you sub (a/e, 0)
 

cutemouse

Account Closed
Joined
Apr 23, 2007
Messages
2,250
Gender
Undisclosed
HSC
N/A
Yeah, but how do I know that it intersects at (a/e, 0)? It could intersect at any point [ie. (a/e, y)]
 

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,178
Gender
Male
HSC
2006
Yeah, but how do I know that it intersects at (a/e, 0)? It could intersect at any point [ie. (a/e, y)]
Read the question:
"The tangent cuts the x-axis at T"
and
"T is the point of intersection between the tangent at point P and one of the directrices of the ellipse"
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top