Conics Question (1 Viewer)

[FD3S-R]

6.
a) Show that if y=mx+k is a tangent to the rectangular hyperbola xy=c^2, then k^2+4mc^2=0

b) Hence find the equations of the tangents from the point (-1, -3) to the rectangular hyperbola xy=4 and find the coordinates of their points of contact.

a) i can do, its just the discriminant.

b) need help!!

 

[KFunk]

6.
a) Show that if y=mx+k is a tangent to the rectangular hyperbola xy=c^2, then k^2+4mc^2=0

b) Hence find the equations of the tangents from the point (-1, -3) to the rectangular hyperbola xy=4 and find the coordinates of their points of contact.

a) i can do, its just the discriminant.

b) need help!!

I havn't done this kind of conics questions for a while but I remember that they really used to piss me off, here goes.

Consider the fact that m = -1/t² [gradient of (cp, c/p) point of contact] so:
mct + c/t = 0 (1)

(cp, c/p) satisfies y = mx + k so:
mct - c/t = -k (2)

(1) + (2) yields 2mct = -k so:
x = -k/2m

(1) - (2) yields 2c/t = k so:
y = k/2

You then have the coordinates of the points of contact in terms of k (-k/2m , k/2)

y = mx + k is satisfied by (-1, -3) and can be written as y + 3 = m(x+1) ---> y = mx + m-3 , ∴ k = m-3

by combining k= m - 3 and k² + 16m = 0 (since c² = 4) we obtain: m² + 10m + 9 = 0 where m = -1 or -9

when m = -1 , k= -4 ..... when m= -9, k= -12

this gives us the point of contact (-2, -2) with the tangent y = -x -4 and the point (-2/3 , -6) with the tangent y = -9x - 12 using y = mx + k and point of contact is (-k/2m, k/2)


There is probably a method which is elegant, this is just how I did it ages ago. If anyone can offer such a method, please do because this is way too long.

 

[KFunk]

I think it's nice to have multiple answers, especially for conics when you have so many different paths you can take.

edit: damn, you should post it back up. Mine isn't that great in the first place.

 

[brett86]

its very similar to yours though...

i think ur solutions very good, its a lot clearer than mine

[EDIT: correct an error in the second tangent]

 

[FD3S-R]

9.
a) Show that if y=mx+k is a tangent to the hyperbola x^2/a^2 - y^2/b^2 =1, then m^2.a^2 - b^2 = k^2

if y = mx+k touches the hyperbola at point p(asecθ, btanθ), then:
m = b/asinθ
amsinθ = b
amsinθ - b = 0 (1)

also P must satisfy y=mx+k ie:
btanθ = amsecθ+k
amsecθ - btanθ = -k (2)

how do i make this into m²a² - b² = k²

thanks

 

[yrtherenonames]

Isn't it the same as the ellipse one?
Sub in y=mx+c into x^2/a^2-y^2/b^2=1 and let delta=0
Although the way you've done it is probably just about there anyway

 

[KFunk]

9.
a) Show that if y=mx+k is a tangent to the hyperbola x^2/a^2 - y^2/b^2 =1, then m^2.a^2 - b^2 = k^2

if y = mx+k touches the hyperbola at point p(asecθ, btanθ), then:
m = b/asinθ
amsinθ = b
amsinθ - b = 0 (1)

also P must satisfy y=mx+k ie:
btanθ = amsecθ+k
amsecθ - btanθ = -k (2)

how do i make this into m²a² - b² = k²

thanks

I think this is your problem: "m = b/asinθ" , --> at P dy/dx = bsecθ/atanθ = m

a.m.tanθ - bsecθ = 0 (1)

as you said: amsecθ - btanθ = -k (2)

(1)² = a²m²tan²θ + b²sec²θ -2ambtanθsecθ = 0

(2)² = a²m²sec²θ + b²tan²θ - 2ambtanθsecθ = k²

(1)² - (2)² =

a²m²(tan²θ - sec²θ ) + b²(sec²θ - tan²theta; ) = -k²

-a²m² + b² = - k²

∴ a²m² - b² = k²

 

[FD3S-R]

yeah i just found out and someone posted reply before i can delete hehe

thanks anyway

 

[lum]

hey kfunk, good solutions!, but i got really lost at this part...
Consider the fact that m = -1/t2 [gradient of (cp, c/p) point of contact] so:
mct + c/t = 0 (1)

well if i sub bak x and y in, it's mx + y = 0, where was this from? and what's with the consider the fact that m = -1/t^2, where'd you use that?

i then did it myself as i had no idea why u did that, couldn't u just start at subbing in the pt into the tangent, get values for m, get values for k and work it out? i mean u didn't really use the first part of the working anyways.(prolly just that i'm not used for finding an expression first in terms of k and m though.., haven't done conics in a long time)

and brett, gw on ur sol'n too, but i don't understand the derivatives part, actually the last 4 lines... i did exactly the same thing as you up to the m=-1, k=-4 and m=-9, k=-12

i'm so failing the bloody cssa on monday...