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If the tangent to the curve 3xy^2-2x^2y=1 at (1,1) intersect the curve again at (alpha,beta) then find (alpha,beta)
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\[\Huge 3xy^2-2x^2y=1\]
find the equation of the line and find the points of intersection.
\[\LARGE 3y^2+6xy \frac{dy}{dx}-(4xy+2x^2 \frac{dy}{dx})=0\] \[\LARGE \frac{dy}{dx}=\frac{4xy-3y^2}{6xy-2x^2}\]
plug in 1,1
4-3/6-2 1/4
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Then use y-1 = m(x-1) to find the line equation
y-1=1/4(x-1)
4y-4=x-1 4y-x-3=0
Yeah, solve the equations.
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