Question

Find out the closest distance of origin to intersect line of the surfaces x+y=z and x*y=1

Answer 1

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Answer 2

ok i am assuming you want the point that minimizes the distance from the origin to the intersection of the 2 surfaces.
the distance of any point from the origin is:
\[d = \sqrt{x^{2}+y^{2}+z^{2}}\]
set 2 surfaces equal to each other
\[x+y = xy -1\]
\[\rightarrow y = \frac{x+1}{x-1}\]
substitute this into surface equation to get z in terms of x
\[z=x+y = x +\frac{x+1}{x-1}\]
\[\rightarrow z =\frac{x^{2}+1}{x-1}\]
\[d = \sqrt{x^{2}+\frac{(x+1)^{2}}{(x-1)^{2}}+\frac{(x^{2}+1)^{2}}{(x-1)^{2}}}\]
We want to minimize d so differentiate and set equal to 0
This can get messy, be careful
simplify radical first, then use quotient rule to differentiate
\[d ^{\prime}= \frac{2(4x ^{3}-3x ^{2}+4x+1)(x-1) - \sqrt{2(x ^{4}-x ^{3}+2x ^{2}+x+1)}}{2(x-1)^{2}\sqrt{2(x ^{4}-x ^{3}+2x ^{2}+x+1)}}=0\]
This simplifies to an 8th degree polynomial, use software to solve this
I used Newtons method and found about 4 real solutions
the one giving the smallest d will be the answer
\[x \approx -0.292\]
\[y \approx -0.548\]
\[z \approx -0.84\]
\[d \approx 1.045\]