help with conics question (1 Viewer)

[polythenepam]

hi
i need help with the last part of this question (question 4 from the 1995 cath trial)
the question goes:

The hyperbola xy=c^2 meets the ellipse x^2/a^2+y^2/b^2=1 at P(ct(1) , c/t(1))
and Q(ct(2) , c/t(2)) (the 1 and the 2 in t(1) and t(2) are just the distinction between the parameters)
where t(1)>t(2)>0.
Tangents to the hyperbola at P and Q meet in T, while tangents to the ellipse at P and Q meet in V.
incase theres important relevance the question asks previously
i)show this information on a sketch
ii) show that the parameter t of a point (ct,c/t) where xy=c^2 and x^2/a^2+y^2/b^2=1 intersect satisfies the equation b^2.c^2.t^4-a^2.b^2.t^2+a^2.c^2=0
iii) show the coordinates of T are ( 2ct(1)t(2)/(t(1)+t(2)) , 2c/(t(1)+t(2)) )
iv) show the coordinates of V are ( a^2/c(t(1)+t(2)) , b^2.t(1).t(2)/c(t(1)+t(2)) )
v) show VT passes through the origin
and then it asks
vi) show that if V lies at a focus of the hyperbola, then the ellipse is in fact a circle and find the radius of this circle in terms of c.

now i did all parts of the question except for the very last which is to find the radius in terms of c..
the answer is r=c(1+5^1/2)^1/2
the furthest i could get was by using b^2.c^2.t^4-a^2.b^2.t^2+a^2.c^2=0
then because b = a = r it becomes c^2.t^4-r^2.t^2+c^2=0
n then by using sum of roots t(1)^2 + t(2)^2 = r^2/c^2
r=c(t(1)^2 + t(2)^2)^1/2
so im assuming we have to prove t(1)^2 + t(2)^2= 1+5^1/2 ?? but i have no idea where this will come from..
if anyone can show how this is found that would be great

 

[KFunk]

[And then I discover that you want the radius, rather than the fact that it is a circle. It would pay to read the question it seems... le sigh] A slightly different method which seems to do it:

The gradient of VT:

m_VT= (b²t₁t₂ - 2c²)/(a² - 2c²t₁t₂)

You've shown that VT passes through (0,0) so

subbing (0,0) into y - y_V = m_VT(x - x_V) yields:

y_V = m_VTx_V

(b²t₁t₂)/c(t₁+t₂) = (b²t₁t₂ - 2c²)/(a² - 2c²t₁t₂) . a²/c(t₁+t₂)

a²b²t₁t₂ - 2b²c²(t₁t₂)² = a²b²t₁t₂ - 2a²c²

(t₁t₂)² = a²/b²

a = bt₁t₂

*phew* , they've told you that V lies at a focus of the hyperbola, i.e (c√2,c√2) so:

c√2 = a²/c(t₁+t₂)
a² = c²√2(t₁+t₂)

and

c√2= (b²t₁t₂)/c(t₁+t₂)
b²t₁t₂ = c²√2(t₁+t₂)

hence

a² = b²t₁t₂

but, a = bt₁t₂

so a² = ab ---> a=b ∴ the ellipse is in fact a circle

There were far too many tags up there so there's bound to be a mistake. Please pull me up on any of them.

 

[polythenepam]

ohh ohk thanx but as u realised i already got this part
lol
but yeeh im pretty sure there are no significant mistakes i double checked it but
if u hav the 95 cath trial then it would help to look at it from there
i was considering that maybe the book im doing it from left out something crucial.. but if not then i must be overlooking something ??

 

[who_loves_maths]

here's an alternative solution to that of KFunk's: (Note: it's not necessarily faster or anything, just an alternative)

the coordinates of 'P' and 'Q' must satisfy that of the ellipse:

ie. (c^2*t(1)^2)/a^2 + c^2/(b^2*t(1)^2) = 1 , and, (c^2*t(2)^2)/a^2 + c^2/(b^2*t(2)^2) = 1

---> t(1)^2/a^2 + 1/b^2*t(1)^2 = 1/c^2 , and, t(2)^2/a^2 + 1/b^2*t(2)^2 = 1/c^2

equating both: t(1)^2/a^2 + 1/b^2*t(1)^2 = t(2)^2/a^2 + 1/b^2*t(2)^2

which reduces to: ---> a^2 = b^2*t(1)^2*t(2)^2 ................................. (1)

'V' is a focus of the hyperbola. the foci of ALL rectangular hyperbola lies on the line y=x , ie. the 'x' and 'y' coordinates are the SAME.
hence, using the coordinates of 'V' found in a previous section you did:

a^2/c(t(1)+t(2)) = b^2.t(1).t(2)/c(t(1)+t(2))

denominators cancel and you get: ---> a^2 = b^2*t(1)*t(2) ................................. (2)

and from the simple equations (1) and (2) , you immediately get:

a^2 = b^2

---> e = 0 ; so the 'ellipse' is infact a circle , with radius 'a' units.


Edit: this will seem more succinct to the eye if i could do subscripts like KFunk did

 

[KFunk]

I'm not that good with my conic properties but if you have a hyperbola like this intersecting a circle then would symmetry dictate that
t₃ = -t₁ and t₄ = -t₂ ?

If that's the case you could use the product of roots to quickly find that t₁t₂ = 1 (from r²c²t⁴ - r⁴t² + r²c²= 0 )

This might then be helpful in finding the radius (and you get it from a=b anyway).

 

[who_loves_maths]

Originaly Posted by KFunk
I'm not that good with my conic properties but if you have a hyperbola like this intersecting a circle then would symmetry dictate that
t3 = -t1 and t4 = -t2 ?

If that's the case you could use the product of roots to quickly find that t1t2 = 1 (from r2c2t4 - r4t2 + r2c2= 0 )

This might then be helpful in finding the radius (and you get it from a=b anyway).

yea, you get that equation from a=b anyways, and you have to prove t(1)*t(2)=1 in the first place to show that a=b...

Originally Posted by polythenepam
now i did all parts of the question except for the very last which is to find the radius in terms of c..
the answer is r=c(1+5^1/2)^1/2
the furthest i could get was by using b^2.c^2.t^4-a^2.b^2.t^2+a^2.c^2=0
then because b = a = r it becomes c^2.t^4-r^2.t^2+c^2=0
n then by using sum of roots t(1)^2 + t(2)^2 = r^2/c^2
r=c(t(1)^2 + t(2)^2)^1/2
so im assuming we have to prove t(1)^2 + t(2)^2= 1+5^1/2 ?? but i have no idea where this will come from..
if anyone can show how this is found that would be great

to find 'a' (or 'r') in terms of 'c', you need to use Parts (iv) and (vi) again polythenepam :

c^2.t^4-r^2.t^2+c^2=0

and as you said: ---> t(1)^2 + t(2)^2 = r^2/c^2 = a^2/c^2 = b^2/c^2 ............ (1)

the coordinates of 'V' is: V(a^2/c(t(1)+t(2)) , b^2.t(1).t(2)/c(t(1)+t(2)) = V(c√2,c√2)

---> a^2/c(t(1)+t(2)) = c√2 ---> (t(1) + t(2)) = a^2/c^2.√2

---> (t(1) + t(2))^2 = t(1)^2 + t(2)^2 +2.t(1).t(2) = t(1)^2 +t(2)^2 + 2 = a^4/2.c^4 ; where t(1).t(2) =1

ie. t(1)^2 + t(2)^2 = a^4/2.c^4 -2 .......... (2)


hence, substitute (2) into (1) to get:

a^4/2.c^4 -2 = a^2/c^2

---> 0 = a^4 -2c^2.a^2 -4c^4 ; which is a quadratic equation in 'a^2'.

solving for 'a^2': a^2 = [2c^2 +/- √(4c^4 + 16c^4)]/2 = c^2.(1 +/- √(5))

---> a = r = c.√(1 +/- √(5))

but in this case, 'a' denotes the radius which is a measure of length. ie. a > 0
so, since c > 0, then 'a' does NOT = c.√(1 - √(5)) < 0

therefore, r = c.√(1 + √(5)) units.


hope that helps

 

[polythenepam]

ohhhhkay i see now..
thanx heaps! that helps very much ur a genius..
lol