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The vertex of the parabola `x^2 + y^2 - 2xy - 4x - 4y + 4 = 0` is atA. (1, 1)B. (-1, -1)C. `(1/2,1/2)`D. none of these |
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Answer» Correct Answer - C we have, `x^(2)+y^(2)-2xy-4x-4y+4=0` `rArr" "(x-y)^(2)=4(x+y-1)` `rArr" "{(x-y)/(sqrt(1^(2)+1^(2)))}^(2)=2sqrt2{(x+y-1)/(sqrt(1^(2)+1^(2)))}` `rArr" "(x-y)/(sqrt(1^(2)+1^(2)))=" Y and "(x+y-1)/(sqrt(1^(2)+1^(2)))=X` Then, equation (i) reduces to `y^(2)=2sqrt2X`. The coordinates of the vertex of this parabola are (X = 0, Y = 0). So, the coordinates of the vertex of the given parabola are given by `(x+y-1)/(sqrt(1^(2)+1^(2)))=" 0 and "(x-y)/(sqrt(1^(2)+1^(2)))=0` i.e. x+y-1=0 and x-y=0 `rArr" "x=y=1/2` Hence, the coordinates of the vertex are (1/2, 1/2). |
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