Another Hyperbola Q (1 Viewer)

[OLDMAN]

The tangent at P(asec@,btan@) on a hyperbola meets the asymptotes at QR. Show that QR is twice the distance of the chord joining point P with the intersection of the asymptotes.

Note: this question is a morph of Geha's question for the special rectangular hyperbola case ie. P(cp,c/p).

 

[ND]

I was only able to prove that this is true for a rectangular hyp:

Q[a/(sec@ - tan@), b/(sec@ - tan@)], R[a/(sec@ + tan@), -b/(sec@ + tan@)] from the other question.

Because P is the midpoint of QR, and O(0,0) is the intersetion of the asymptotes, for this to be true:

OP = PQ
OP^2 = PQ^2

OP^2 = a^2*(sec@)^2 + b^2*(tan@)^2

PQ^2 = [a/(sec@-tan@) - asec@]^2 + [b/(sec@-tan@) - btan@]^2
= [a(1 - sec@(sec@-tan@))/(sec@-tan@)]^2 + [b(1 - tan@(sec@-tan@))/sec@-tan@)]^2
= [atan@(sec@-tan@)/(sec@-tan@)]^2 + [bsec@(sec@-tan@)/(sec@-tan@)]^2
= a^2*(tan@)^2 + b^2*(sec@)^2
= a^2*(sec@)^2 + b^2*(tan@)^2 - (a^2 - b^2)
= OP^2 - (a^2 - b^2)
.'. statement is only true when a=b (i.e. hyperbola is rectangular).

 

[OLDMAN]

ND: you had me worried, so I rechecked my calculations. The area of hyperbola triangle question was also morphed out of Geha's who assumes the easier rectangular hyperbola case "area of triangle property"; that's why I became suspicious of this "length of intercept property", whether it could apply to any hyperbola. And yes it does. So a note for all Geha book lovers : these two properties also apply to the non-rectangular hyperbola as well.

Re. your computations: why have you assumed P is the midpoint of QR? why are you computing PQ?
The question boils down to prove QR=2*OP.

 

[underthesun]

seems like, when I use the distance equation i got RQ as ::

X distance ::
= a/(sec@ - tan@) + a/(sec@ + tan@)
= (asec@ + atan@ - asec@ + atan@)/(sec^2@ - tan^2@)
= 2atan@

similarly,
Y distance = 2asec@

RQ = sqrt(X^2 + Y^2);

RQ = sqrt(4*a^2 tan^2 @ + 4*b^2 sec^2 @);

RQ = 2 * sqrt(a^2 tan^2 @ + b^2 sec^2 @);

while you know the distance OP..

which means that its only true when a = b, which is rectangular hyperbola, but again im never sure..

 

[ND]

Originally posted by OLDMAN

Re. your computations: why have you assumed P is the midpoint of QR? why are you computing PQ?
The question boils down to prove QR=2*OP.

Isn't it a property of a hyperbola that for a tangent at P that cuts the asymptotes at Q and R, the midpoint of QR is P?
Heres a proof:

Q[a/(sec@ - tan@), b/(sec@ - tan@)], R[a/(sec@ + tan@), -b/(sec@ + tan@)]

MP_x = a/2(sec@-tan@) + a/2(sec@+tan@)
= [a(sec@+tan@) + a(sec@-tan@)]/2
= asec@
same thing happens for y = btan@.
.'. P is the midpoint of QR.

The reason i was using PQ is because PQ = QR/2. So essentially i was trying to prove QR = 2OP.

 

[OLDMAN]

Had problems logging into the site.

I also beg your pardon, thought this "length of intercept property" generalized as did the "area of triangle property", got my atan and bsec crossed, I guess we can't nail Geha twice.

assume s=sec@, t=tan@

OP^2=(as)^2 + (bt)^2
QR^2=(a/(s-t)-a/(s+t))^2+(b/(s-t)+b/(s+t))^2
=(2at)^2+(2bs)^2=4((at)^2+(bs)^2))

OP=QR/2 for all @ only when a=b rectangular.

Interesting property. ND Q=QR/2 well done. Underthesun: good work as well!