9ah u r giving unneceesary info to confuse us ;)
k the proof has noting to do wit line
wen asymptotes intersect they are symmetrical and use SAS congruency to get result
I know many of you would've done this problem before still i thought it would be fun to check your presence of mind(not knowledge in coordinate!!!!)
Assume an arbitary st. line cutting a hyperbola (assume x2/a2-y2/b2=1) and its asymptotes at 4 distinct points. prove length of two small segments,created by interception of hyperbola and asympt. , between hyperbola and asymptotes are equal..
LINE LIMIT: 4 LINES.
TIME: 1.5 MINUTES
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UP 0 DOWN 0 1 7
7 Answers
91hey the asymptotes here are an arbitrary line!!
they are not parallel to the y axis!!
1i will give you the figure. its not symmetric ;)
consider any line y=mx+c
1just put y=mx+c in the equation of asymptotes and hyperbola that are
x2/a2-y2/b2=0 and x2/a2-y2/b2=1 .we see that sum of roots are same in both that means the are bisected at same point hence intercepts b/w hyperbola and asym are equal.
1too dirty,, BUT VERY GOOD WAY OF SOLVING,
((I was thinking of a similar one. :):) Here it goes----------))
assume intersections A,B,C,D.
take midpt. of AD=(h,k)
apply T=S1 for (i)hyperbola; for(ii) pair of st. lines(asymptotes,they are conic too!! eqn. =just throw off the 1 in hyp. eqn.)
same result for midpt. of BC. hence AB=CD..
1i never said you dont use it, i said dont use it in deep for dirty calculation anyway i just wrote equations, no need to solve them :)