2020 H2 Mathematics Paper 1 Question 2

Differentiation I: Tangents and Normals, Parametric Curves

Answers

5x+9y=14{5x+9y=14}

Full solutions

x21+x2+y21+y2=x3y51−11+x2+1−11+y2=x3y5\begin{gather*} \frac{x^2}{1+x^2} + \frac{y^2}{1+y^2} = x^3 y^5 \\ 1 - \frac{1}{1+x^2} + 1 - \frac{1}{1+y^2} = x^3 y^5 \\ \end{gather*}
Differentiating w.r.t. x,{x,}
2x(1+x2)2+2y(1+y2)2dydx=3x2y5+5x3y4dydx\frac{2x}{(1+x^2)^2} + \frac{2y}{(1+y^2)^2}\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 y^5 + 5x^3y^4 \frac{\mathrm{d}y}{\mathrm{d}x}
At (1,1),{(1,1), }
2(1+1)2+2(1+1)2dydx=3+5dydx92dydx=−52dydx=−59\begin{gather*} \frac{2}{(1+1)^2} + \frac{2}{(1+1)^2} \frac{\mathrm{d}y}{\mathrm{d}x} = 3 + 5 \frac{\mathrm{d}y}{\mathrm{d}x} \\ \frac{9}{2}\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{5}{2} \\ \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{5}{9} \end{gather*}
Equation of tangent at (1,1):{(1,1):}
y−1=−59(x−1)9y−9=−5x+55x+9y=14  ■\begin{gather*} y - 1 = -\frac{5}{9} (x-1) \\ 9y - 9 = -5x+5 \\ 5x + 9y = 14 \; \blacksquare \\ \end{gather*}