A curve has equation \(\frac{x^2}{1+x^2}+\frac{y^2}{1+y^2}=x^3y^5\). Find the equation of the tangent to the curve at the point \((1,1)\). Give your answer in the form \(ax+by=c\), where \(a\), \(b\) and \(c\) are integers.[6]
Implicit differentiation gives \(\frac{2x}{(1+x^2)^2}+\frac{2y}{(1+y^2)^2}\frac{\mathrm dy}{\mathrm dx}=3x^2y^5+5x^3y^4\frac{\mathrm dy}{\mathrm dx}\).
At \((1,1)\), \(\frac12+\frac12 y'=3+5y'\), so \(y'=-\frac59\).
The tangent is \(y-1=-\frac59(x-1)\), hence \(5x+9y=14\).
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