According to the definition of reciprocal cone, we want to find the reciprocal of
x2−y2+5z2+4xy=0
Equation of reciprocal of x2−y2+5z2+4xy=0 is;
Ax2+By2+Cz2+2Fyz+2Gxz+2Hxy=0
where,
A=bc−f2, B=ac−g2, C=ab−h2F=gh−af, G=hf−bg, H=fg−ch
Here, a=1, b=−1, c=5, f=0, g=0, h=2.
So,
A=bc−f2=−5−0=−5, B=ac−g2=5−0=5, C=ab−h2=−1−4=−5F=gh−af=0−0=0, G=hf−bg=0−0=0, H=fg−ch=0−10=−10
Thus, equation of reciprocal of x2−y2+5z2+4xy=0 is;
−5x2+5y2−5z2+(2×0×yz)+(2×0×xz)−20xy=0⇒−5x2+5y2−5z2−20xy=0
Dividing through by -5;
⇒x2−y2+z2+4xy=0, which is the other cone given in the question.
Showing that that the perpendiculars drawn from the origin to tangent planes to the cone
x2−y2+5z2+4xy=0 lie on the cone x2−y2+z2+4xy=0 .
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