The equation 13[{(x - 1)^2} + {(y - 2)^2}] = | vedclass

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(C) The given equation is $13[(x - 1)^2 + (y - 2)^2] = 3(2x + 3y - 2)^2$.This is of the form $SP^2 = e^2 PM^2$,where $S$ is the focus,$P$ is a point $

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Hyperbola

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(C) The given equation is $13[(x - 1)^2 + (y - 2)^2] = 3(2x + 3y - 2)^2$.This is of the form $SP^2 = e^2 PM^2$,where $S$ is the focus,$P$ is a point $(x, y)$,and $PM$ is the perpendicular distance from $P$ to the directrix.Expanding the equation:$13(x^2 - 2x + 1 + y^2 - 4y + 4) = 3(4x^2 + 9y^2 + 4 + 12xy - 8x - 12y)$$13x^2 + 13y^2 - 26x - 52y + 65 = 12x^2 + 27y^2 + 36xy - 24x - 36y + 12$$x^2 - 36xy - 14y^2 - 2x - 16y + 53 = 0$.Comparing this with the general second-degree equation $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$,we have $A = 1$,$H = -18$,and $B = -14$.The discriminant $H^2 - AB = (-18)^2 - (1)(-14) = 324 + 14 = 338$.Since $H^2 - AB > 0$,the conic section represents a hyperbola.

The equation $13[{(x - 1)^2} + {(y - 2)^2}] = 3{(2x + 3y - 2)^2}$ represents

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