Conic Hyperbola (1 Viewer)

[OLDMAN]

Prove that the area of the triangle formed by the tangent to the hyperbola and the asymptotes is a constant.

 

[ND]

Ah my teacher used this question as an example when we did hyperbolas.

let A and B be the points of intersection of tangent and the asymptotes.

eqn of tangent to hyperbola at P(asec@, btan@):

xsec@/a - ytan@/b = 1 ...[1]

eqn's of asymptotes:

y = +-bx/a

sub y = bx/a into [1]:

xsec@/a - atan@/a = 1
x[(sec@ - tan@)/a] = 1
x = a/(sec@ - tan@)

similarly upon subbing x = ay/b into [1]:

y = b/(sex@ - tan@)

.'. A[a/(sec@ - tan@), b/(sec@ - tan@)]

subbing y = -bx/a and x = -ay/b gets B[a/(sec@ + tan@), -b/(sec@ + tan@)]

let C be the point of intersection of the tengent with the x axis.

subbing y = 0 into [1] gives:

x = a/sec@

.'. C(a/sec@, 0)

Now area of /\OAB = area /\OAC + area /\OBC

area /\OAC = ab/2sec@(sec@ - tan@) (using 1/2*base*height rule)
area /\ABC = ab/2sec@(sec@ + tan@)

so area /\OAB = ab/2sec@(sec@ - tan@) + ab/2sec@(sec@ - tan@)
= ab/2sec@[1/(sec@ - tan@) + 1/(sec@ + tan@)]
= ab/2sec@[2sec@/((sec@)^2 - (tan@)^2)]
= ab/((sec@)^2 - (tan@)^2)
= ab

 

[OLDMAN]

Great. Splitting that triangle so as to have the same horizontal base is definitely elegant!

 

[underthesun]

hey.., in exam situations, are you required to deduce the tangent equations?

and, it seems that my way requires using the distance formula..

is there a better method than this?

edit: I see the answer now. seems i posted a bit too late..

edit: indeed elegant

 

[OLDMAN]

The question was not posted like an exam question. If this q was in the exam, there would be a few leading questions- it is amazing how the examiner leads you by the hand. Perhaps part i) might be : show that eqn. of tangent is xsec@/a-ytan@/@=1;
part ii) show equation of the asymptotes;
part iii) might be: hence find that the area of the triangle formed by the tangent, the asymptote and the x axis is

 

[underthesun]

oh i see.. thanks.

 

[OLDMAN]

Underthesun :"and, it seems that my way requires using the distance formula.."

Not all is lost, I am posting another Hyperbola problem in a new thread using the distance formula.