haboozin
Do you uhh.. Yahoo?
- Joined
- Aug 3, 2004
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- HSC
- 2005
Conics Help Marathon
<b>QUESTION 1:</b>
P(acos@,bsin@) lies on the ellipse x<sup>2</sup>/a<sup>2</sup> + y<sup>2</sup>/b<sup>2</sup> = 1, where a > b > 0. The tanget to the ellipse at P passes through a focus of the hyperbola x<sup>2</sup>/a<sup>2</sup> - y<sup>2</sup>/b<sup>2</sup> = 1 with eccentricity e.
i, Show that the tangent to the ellipse at P has equation bxcos@ + aysin@ = ab
(i think thats a typo cos i get bxcos@ + aysin@ = ae)
ii,
show that P lies on a directrix of the hyperbola.
iii,
show that the tangent to the ellipse at P has gradient +-1
<b>QUESTION2:</b>
ok i know this has something to do with relationships, but i forgot these relationships..
P(acosA,bsonA) and Q(acosB,bsinB) lie on the ellipse x<sup>2</sup>/a<sup>2</sup> + y<sup>2</sup>/b<sup>2</sup> = 1
i,
verify that the coordinates of P satisfy x/a cos((A + B)/2) + y/b sin((A + B)/2) = cos((A - B)/2)
Note: A and B are angles and have nothing to do with a and b.
ii, deduce that the chord PQ has equation x/a cos((A + B)/2) + y/b sin((A + B)/2) = cos((A - B)/2)
this is very easy really, because if p works in there so willl q therefore it willl be a chord..
iii, hence or otherwise show that if PQ is a focal chord of the ellipse, then cos((A-B)/2) = +-ecos((A + B)/2)
<b>QUESTION3:</b>
The line l is a common tangent to the hyperbolas xy=cb<sup>2</sup>, xb<sup>2</sup>/ab<sup>2</sup> - yb<sup>2</sup>/bb<sup>2</sup>=1 with points of contact P (ct, c/t) and Q (asec@,btan@) respectively
you may assume: tangent at P is x +t<sup>2</sup>=2ct and tangent at Q is xsec@/a - ytan@/b = 1
deduce that:
sec@/a = -tan@/bt^2 = 1/2ct
<b>QUESTION 4:</b>
P (asec@,btan@) lies on the hyperbola x<sup>2</sup>/a<sup>2</sup> - y<sup>2</sup>/b<sup>2</sup> = 1. the tangent at P is also a tangent of a circle. (x - ae)<sup>2</sup> + y<sup>2</sup> = a<sup>2</sup>(e<sup>2</sup>+1)
show that sec@ = -e
Ok, for this question first i tried to solve the equation of the tangent of the hyperbola simultaniously with the circle when DELTA = 0. That came out to be too long! so then i tried to implicietly differentiate both the circle and the hyperbola then put in P in both of then and solve them when they both equal and i got sec@ = 1/e . I then tried to use how tangents to the circle are always perpendicular to the center and still got sec@ = 1/e
btw, this was in our term test today.
If you are doing these, full working will be appreciated dont just tell me what to do because i already have at least some idea of how to do these questions and i really need to see the proofs.
<b>QUESTION 1:</b>
P(acos@,bsin@) lies on the ellipse x<sup>2</sup>/a<sup>2</sup> + y<sup>2</sup>/b<sup>2</sup> = 1, where a > b > 0. The tanget to the ellipse at P passes through a focus of the hyperbola x<sup>2</sup>/a<sup>2</sup> - y<sup>2</sup>/b<sup>2</sup> = 1 with eccentricity e.
i, Show that the tangent to the ellipse at P has equation bxcos@ + aysin@ = ab
(i think thats a typo cos i get bxcos@ + aysin@ = ae)
ii,
show that P lies on a directrix of the hyperbola.
iii,
show that the tangent to the ellipse at P has gradient +-1
<b>QUESTION2:</b>
ok i know this has something to do with relationships, but i forgot these relationships..
P(acosA,bsonA) and Q(acosB,bsinB) lie on the ellipse x<sup>2</sup>/a<sup>2</sup> + y<sup>2</sup>/b<sup>2</sup> = 1
i,
verify that the coordinates of P satisfy x/a cos((A + B)/2) + y/b sin((A + B)/2) = cos((A - B)/2)
Note: A and B are angles and have nothing to do with a and b.
ii, deduce that the chord PQ has equation x/a cos((A + B)/2) + y/b sin((A + B)/2) = cos((A - B)/2)
this is very easy really, because if p works in there so willl q therefore it willl be a chord..
iii, hence or otherwise show that if PQ is a focal chord of the ellipse, then cos((A-B)/2) = +-ecos((A + B)/2)
<b>QUESTION3:</b>
The line l is a common tangent to the hyperbolas xy=cb<sup>2</sup>, xb<sup>2</sup>/ab<sup>2</sup> - yb<sup>2</sup>/bb<sup>2</sup>=1 with points of contact P (ct, c/t) and Q (asec@,btan@) respectively
you may assume: tangent at P is x +t<sup>2</sup>=2ct and tangent at Q is xsec@/a - ytan@/b = 1
deduce that:
sec@/a = -tan@/bt^2 = 1/2ct
<b>QUESTION 4:</b>
P (asec@,btan@) lies on the hyperbola x<sup>2</sup>/a<sup>2</sup> - y<sup>2</sup>/b<sup>2</sup> = 1. the tangent at P is also a tangent of a circle. (x - ae)<sup>2</sup> + y<sup>2</sup> = a<sup>2</sup>(e<sup>2</sup>+1)
show that sec@ = -e
Ok, for this question first i tried to solve the equation of the tangent of the hyperbola simultaniously with the circle when DELTA = 0. That came out to be too long! so then i tried to implicietly differentiate both the circle and the hyperbola then put in P in both of then and solve them when they both equal and i got sec@ = 1/e . I then tried to use how tangents to the circle are always perpendicular to the center and still got sec@ = 1/e
btw, this was in our term test today.
If you are doing these, full working will be appreciated dont just tell me what to do because i already have at least some idea of how to do these questions and i really need to see the proofs.
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