A) \[h+h'=0\]
B) \[h=h'\]
C) \[h+h'=1\]
D) None of these
Correct Answer: A
Solution :
\[S\equiv (si{{n}^{2}}\theta )\,\,{{x}^{2}}+2hxy+({{\cos }^{2}}\theta )\,\,{{y}^{2}}+32x+16y+19=0\] | ??...(1) |
\[S'\equiv ({{\cos }^{2}}\theta )\,\,{{x}^{2}}+2h'xy+({{\sin }^{2}}\theta )\,\,{{y}^{2}}+16x+32y+19=0\] | ???(2) |
\[\because \]Curve passing through point of intersection of S & S? is \[S+\lambda S'=0\] | |
\[\Rightarrow {{x}^{2}}({{\sin }^{2}}\theta +\lambda {{\cos }^{2}}\theta )+{{y}^{2}}({{\cos }^{2}}\theta +\lambda {{\sin }^{2}}\theta )+\] | |
\[2xy\,\,(h+\lambda h')+x\,\,(32+16\lambda )+y\,\,(16+32\lambda )+19\,\,(1+\lambda )=0\]For this equation to be a circle, then | |
\[{{\sin }^{2}}\theta +\lambda {{\cos }^{2}}\theta ={{\cos }^{2}}\theta +\lambda {{\sin }^{2}}\theta \]\[\Rightarrow \lambda =1\] and \[h+\lambda h'=0\]\[\Rightarrow h+h'=0\] |
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