Prove that the locus of poles of tangents to conic ax 2 +2hxy+by 2 =1 with respect to the conic a ' x 2 +b ' y 2 +2hxy=1 is the conic a(h ' x+b ' y) 2 -2h(a ' x+h ' y)(h ' x+b ' y)+b(a ' x+h ' y) 2 =ab-h 2 .

Prove that the locus of poles of tangents to conic ax2+2hxy+by2=1 with respect to the conic a'x2+b'y2+2hxy=1 is the conic a(h'x+b'y)2-2h(a'x+h'y)(h'x+b'y)+b(a'x+h'y)2=ab-h2.

Grade:12th pass

1 Answers

Aditya Gupta
2086 Points
4 years ago
hello dipika, so lemme tell you this is an EXTREMELY lengthy ques. it took me full 6 pages to complete the ques. kamartod ques h yar xdd.
first of all assume a random point (m, n) on ax2+2hxy+by2=1, and write the eqn of tangent at this point to the curve (T=0 in the standard form). since (m, n) lies on ax2+2hxy+by2=1, so am2+2hmn+bn2=1.........(1)
now, find the pole of this line T= 0 wrt the conic a'x2+b'y2+2h’xy=1. to do this, we assume the point whose locus is to be found to be P(u, v). now, we first write the eqn of polar of P wrt a'x2+b'y2+2h’xy=1 using the std result T= 0. but since P is the pole of T= 0, hence T is the polar of P. so, we compare the eqns T= 0 and T= 0 (since they both represent one and the same line, namely the polar of P). by doing so, you shall obtain two eqns in m and n. so, simply solve these 2 eqns for m and n, and then replace those values in (1).
now finally replace u by x and v by y.
however, the eqn you have now is extremely complicated, and it would almost take an eternity to rearrange, cancel and factor out the eqn to its reqd form given above as a(h'x+b'y)2-2h(a'x+h'y)(h'x+b'y)+b(a'x+h'y)2=ab-h2. however, if you try hard enough, you will eventually be able to do it. bas mehnat bahut lagegi. i spent around half an hour on the ques, but in the end the result matches precisely! 
kindly approve :)

Think You Can Provide A Better Answer ?


Get your questions answered by the expert for free