A curve has equation x^2 +2xy–3y^2 +16=0. Find the coordinates of the points on the curve where dy/dx = 0.
x2 +2xy–3y2+16=0Differentiate the terms:x2 gives 2x2xy is differentiated by the product rule: vu' +v'u Make v = 2x and u = y, which gives 2x(dy/dx) + 2y3y2 gives 6y(dy/dx)16 gives 0.Therefore we have the equation: 2x + 2x(dy/dx) + 2y - 6y(dy/dx) = 0Now collect like terms: 6y(dy/dx) - 2x(dy/dx) = 2x + 2yFactor out dy/dx: dy/dx = (2x+2y)/(6y-2x)The question asks for points where dy/dx = 0, so substituting this in gives: 0 = (2x+2y)/(6y-2x) 0 = 2x + 2y y = -xNow to find the points on the curve where dy/dx is 0, we know y=-x and so substituting this into the original equation gives: x2 + 2x(-x) - 3(-x)2 + 16 = 0 x2 - 2x2 - 3x2 + 16 = 0 4x2 = 16 x2 = 4 x = (+/-) 2 Therefore y = (-/+) 2And now we have our answer, the coordinates are (2,-2) and (-2,2)
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