Use implicit differentiation to find an expression for $\frac{{{\text{d}}y}}{{{\text{d}}x}}$.

Find the equation of the normal to the curve at the point (1, 1).

METHOD 1

expanding the brackets first:

${x^4} + 2{x^2}{y^2} + {y^4} = 4x{y^2}$     M1

$4{x^3} + 4x{y^2} + 4{x^2}y\frac{{{\text{d}}y}}{{{\text{d}}x}} + 4{y^3}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 4{y^2} + 8xy\frac{{{\text{d}}y}}{{{\text{d}}x}}$     M1A1A1

Note:     Award M1 for an attempt at implicit differentiation.

     Award A1 for each side correct.

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ - {x^3} - x{y^2} + {y^2}}}{{x{y^2} - 2xy + {y^3}}}$ or equivalent     A1

METHOD 2

$2\left( {{x^2} + {y^2}} \right)\left( {2x + 2y\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right) = 4{y^2} + 8xy\frac{{{\text{d}}y}}{{{\text{d}}x}}$     M1A1A1

Note:     Award M1 for an attempt at implicit differentiation.

     Award A1 for each side correct.

$\left( {{x^2} + {y^2}} \right)\left( {x + y\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right) = {y^2} + 2xy\frac{{{\text{d}}y}}{{{\text{d}}x}}$

${x^3} + {x^2}y\frac{{{\text{d}}y}}{{{\text{d}}x}} + {y^2}x + {y^3}\frac{{{\text{d}}y}}{{{\text{d}}x}} = {y^2} + 2xy\frac{{{\text{d}}y}}{{{\text{d}}x}}$     M1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ - {x^3} - x{y^2} + {y^2}}}{{y{x^2} - 2xy + {y^3}}}$ or equivalent     A1

[5 marks]

METHOD 1

at (1, 1), $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ is undefined     M1A1

$y = 1$     A1

METHOD 2

gradient of normal $=  - \frac{1}{{\frac{{{\text{d}}y}}{{{\text{d}}x}}}} =  - \frac{{\left( {y{x^2} - 2xy + {y^3}} \right)}}{{\left( { - {x^3} - x{y^2} + {y^2}} \right)}}$     M1

at (1, 1) gradient $= 0$     A1

$y = 1$     A1

[3 marks]