The equation 14x^2 - 4xy + 11y^2 - 44x - 58y | vedclass

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(B) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. Comparing this with the given equation,we have $a = 14$

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An ellipse

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(B) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. Comparing this with the given equation,we have $a = 14$,$h = -2$,$b = 11$,$g = -22$,$f = -29$,and $c = 71$. First,we calculate the discriminant $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2$: $\Delta = (14)(11)(71) + 2(-29)(-22)(-2) - 14(-29)^2 - 11(-22)^2 - 71(-2)^2$. $\Delta = 10934 - 2552 - 11774 - 5324 - 284 = -9000 
eq 0$. Since $\Delta 
eq 0$,the equation represents a conic section. Next,we check the condition $h^2 - ab$: $h^2 - ab = (-2)^2 - (14)(11) = 4 - 154 = -150$. Since $h^2 - ab < 0$,the conic section is an ellipse.

The equation $14x^2 - 4xy + 11y^2 - 44x - 58y + 71 = 0$ represents

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