x^2 + xy + 3y^2 = 5 Write the equation of the tangent line at (-2,1) showing all

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x^2 + 3y^2 = 1 + 3xy..........(1) a) x=1 ==> 1^2+3y^2=1+3*1*y ==> y^2=y ==> y^2-y=0 ==> y(y-1)=0 ==> y1=0 and y2=1 There are 2 points: A(1,0) and B(1,1). First derivative y': 2x+6yy'=3y+3xy' y'(6y-3x)=3y-2x ==> y'=(3y-2x)/(6y-3x) 1) At the point A: y'=-2/(-3)=2/3 The equation of the tangent line through point A is: y-0=(2/3)*(x-1) or y=2x/3-2/3 2) At point B: y'=(3-2)/(6-3)=1/3 Tangent line through B: y-1=(1/3)*(x-1) or y=x/3+2/3 b) If the tangent line is vertical, then y'=8 ==> (3y-2x)/(6y-3x)=8 ==> 6y-3x=0 ==> x=2y Substitute x in (1) ==> (2y)^2+3y^2=1+3*2y*y ==> y^2=1 ==> y=-1 (x=-2) or y=1 (x=2)==> The tangent lines are vertical at the points (-2,-1) and (2,1). Their equations are x=-2 and x=2 respectively.

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