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Solution 4.1:7d

From Förberedande kurs i matematik 1

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We rewrite the equation in standard form by completing the square for the x- and y-terms,

x22xy2+2y=(x1)212=(y+1)212.

Now, the equation is

(x1)21+(y+1)21(x1)2+(y+1)2=2=0.

The only point which satisfies this equation is (xy)=(11) because, for all other values of x and y, the left-hand side is strictly positive and therefore not zero.


Image:4_1_7_d.gif