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12 Jul 2007 17:54:23 IST
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prove that the eq. (a+2h+b)x2 -2(a-b)xy +(a-2h+b)y2 =0 represents a pair of lines each inclined at an angle of 45 to one or the other of the lines given by ax2 + 2hxy + by2 =0.
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12 Jul 2007 18:28:33 IST
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IF THE PAIR OF ST. LINES P1 : (a+2h+b)x2 -2(a-b)xy +(a-2h+b)y2 =0 ARE INCLINED AT AN ANGLE 45 DEGREES TO P2 : ax2 + 2hxy + by2 =0 ,THEN P2 MUST BE THE ANGLE BISECTER OF LINES OF P1.
THEREFORE, FIND THE EQUATION OF THE ANGLE BISECTORS OF P1,
YOU WILL GET THE ANSWER .
RATE ME IF U LIKE THE APPROACH
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17 Jul 2007 18:39:10 IST
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no 110391...or watever ur name is.....
the pairs of lines need not be bisectors of their angles.
suppose the first two lines make angle 30deg.
then if the 2nd pair makes 60 deg with each other....the condition is still satisfied....so they need not be bisectors.
i suggest you use m1+m2=-2h/a and m1m2=a/b and use tan A=(m1-m2)/1+m1m2
rate me if im correct
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ameya s,
1 yr
iit bombay.
if you ever want to help me clean my room, go STUDY NOW!!! |
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