Question #293619

Show 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. 


1
Expert's answer
2022-02-10T13:06:07-0500

According to the definition of reciprocal cone, we want to find the reciprocal of

𝑥2𝑦2+5z2+4𝑥𝑦=0𝑥^2 − 𝑦^2 + 5z^2 + 4𝑥𝑦 = 0

Equation of reciprocal of 𝑥2𝑦2+5z2+4𝑥𝑦=0𝑥^2 − 𝑦^2 + 5z^2 + 4𝑥𝑦 = 0 is;

Ax2+By2+Cz2+2Fyz+2Gxz+2Hxy=0Ax^2+By^2+Cz^2+2Fyz+2Gxz+2Hxy=0

where,

A=bcf2, B=acg2, C=abh2F=ghaf, G=hfbg, H=fgchA=bc-f^2,\ B=ac-g^2,\ C=ab-h^2\\ F=gh-af,\ G=hf-bg,\ H=fg-ch


Here, a=1, b=1, c=5, f=0, g=0, h=2a=1,\ b=-1,\ c=5,\ f=0,\ g=0,\ h=2.

So,

A=bcf2=50=5, B=acg2=50=5, C=abh2=14=5F=ghaf=00=0, G=hfbg=00=0, H=fgch=010=10A=bc-f^2=-5-0=-5,\ B=ac-g^2=5-0=5,\ C=ab-h^2=-1-4=-5\\ F=gh-af=0-0=0,\ G=hf-bg=0-0=0,\ H=fg-ch=0-10=-10

Thus, equation of reciprocal of 𝑥2𝑦2+5z2+4𝑥𝑦=0𝑥^2 − 𝑦^2 + 5z^2 + 4𝑥𝑦 = 0 is;

5x2+5y25z2+(2×0×yz)+(2×0×xz)20xy=05x2+5y25z220xy=0-5x^2+5y^2-5z^2+(2\times0\times yz)+(2\times0\times xz)-20xy=0\\ \Rightarrow -5x^2+5y^2-5z^2-20xy=0\\

Dividing through by -5;

x2y2+z2+4xy=0, which is the other cone given in the question.\displaystyle \Rightarrow x^2-y^2+z^2+4xy=0,\ \text{which is the other cone given in the question.}

Showing that that the perpendiculars drawn from the origin to tangent planes to the cone

𝑥2𝑦2+5z2+4𝑥𝑦=0𝑥^2 − 𝑦^2 + 5z^2 + 4𝑥𝑦 = 0 lie on the cone 𝑥2𝑦2+𝑧2+4𝑥𝑦=0𝑥^2 − 𝑦^2 + 𝑧^2 + 4𝑥𝑦 = 0 .


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