Hyperbola question (1 Viewer)

[aud]

Coroneos asks this a lot, and I thought I had it, went back and realised that I'd made it up

The tangent at P (a sec , b tan ) on the hyperbola H: x/a - y/b = 1 meets a directrix in Q and S is the corresponding focus. Prove that SQ is perpendicular to SP.

So you find tangent eqn, find coords of Q, find grad SP, find grad SQ, times them together and you should get -1, right?

I keep ending up with:
gradPS . gradSQ = -b/(a(e-1))... it'd work if that a was an a, but I can't work out where it disappeared...

 

[xiao1985]

Originally posted by aud
The tangent at P (a sec , b tan ) on the hyperbola H: x/a - y/b = 1 meets a directrix in Q and S is the corresponding focus. Prove that SQ is perpendicular to SP.

uhm, this line seems a bit confusin.... r u sure u typed it rite??i dun get the logic of this sentence... tgt meets a directrix in q and x is the correspondin focus... ???

 

[aud]

Originally posted by xiao1985
uhm, this line seems a bit confusin.... r u sure u typed it rite??i dun get the logic of this sentence... tgt meets a directrix in q and x is the correspondin focus... ???

No... the tangent at P cuts the directrix at Q, and the corresponding focus is S... it's typed right

 

[Affinity]

Bashing does work.. think there's a smart way which I forgot.

gradient of tangent = b/[asin(t)]

using that to solve for Q you should get:

Q=(a/e, u)

where u = b[ tan(t) - cosec(t)*(sec(t) - 1/e)]

gradient(PQ)*gradient(SQ)

={ [btan(t) - u]/[asec(t) - a/e] } * { u/[ a/e - ae] }

={b^2 [tan(t) - csc(t)*(sec(t) - 1/e)]*csc(t)(sec(t) - 1/e)} / {a^2 (1/e - e)(sec(t) - 1/e)}

= (b^2/a^2) * [csc(t)*(tan(t) - sin(t)*(sec(t)-1/e)]/(1/e - e)
=(b^2/a^2)*[1/(1-e^2)]

and 1-e^2 = -(b^2/a^2)
so the product is -1

Maybe one of the mods could post something more elegant

 

[aud]

Originally posted by Affinity
Bashing does work.. think there's a smart way which I forgot.

gradient of tangent = b/[asin(t)]

using that to solve for Q you should get:

Q=(a/e, u)

where u = b[ tan(t) - cosec(t)*(sec(t) - 1/e)]

gradient(PQ)*gradient(SQ)

={ [btan(t) - u]/[asec(t) - a/e] } * { u/[ a/e - ae] }

={b^2 [tan(t) - csc(t)*(sec(t) - 1/e)]*csc(t)(sec(t) - 1/e)} / {a^2 (1/e - e)(sec(t) - 1/e)}

= (b^2/a^2) * [csc(t)*(tan(t) - sin(t)*(sec(t)-1/e)]/(1/e - e)
=(b^2/a^2)*[1/(1-e^2)]

and 1-e^2 = -(b^2/a^2)
so the product is -1

Maybe one of the mods could post something more elegant

Woah, you've got almost all the trig ratios there

I did this, can someone see where the mistake is?

x/a - y/b = 1
2x/a - 2y/b dy/dx = 0
dy/dx = bx/ay
at P, dy/dy = bsec/asec

tangent at P:
y - btan = (bsec/atan)(x - asec)
xsec/a - ytan/b = 1

at x = a/e,
sec/e - ytan/b = 1
ytan = bsec/e - b
y = (bsec-be)/etan
tf. Q = (a/e , (bsec-be)/etan )

m(PS) = btan/(asec - ae)
m(SQ) = (bsec-be/etan)/(a/e - ae)
= (bsec-be)/etan * e/(a-ae)
= (bsec-be)/((a-ae)tan)

m(PS)m(SQ) = btan/(asec - ae) * (bsec-be)/((a-ae)tan)
= (b(sec-e)) / (a(e - sec)(e-1))
= - b/(a(e-1))

So an a has disappeared somewhere...

 

[Affinity]

error made on second last line