Find an expression for \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) in terms of \(x\) and \(y\).

Find the equations of the tangents to this curve at the points where the curve intersects the line \(x = 1\).

attempt to differentiate implicitly     M1

\(3 - \left( {4y\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2{y^2}} \right){{\text{e}}^{x - 1}} = 0\)    A1A1A1

Note: Award A1 for correctly differentiating each term.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3 \bullet {{\text{e}}^{1 - x}} - 2{y^2}}}{{4y}}\)    A1

Note: This final answer may be expressed in a number of different ways.

[5 marks]

\(3 - 2{y^2} = 2 \Rightarrow {y^2} = \frac{1}{2} \Rightarrow y =  \pm \sqrt {\frac{1}{2}} \)    A1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3 - 2 \bullet \frac{1}{2}}}{{ \pm 4\sqrt {\frac{1}{2}} }} =  \pm \frac{{\sqrt 2 }}{2}\)    M1

at \(\left( {1,{\text{ }}\sqrt {\frac{1}{2}} } \right)\) the tangent is \(y - \sqrt {\frac{1}{2}}  = \frac{{\sqrt 2 }}{2}(x - 1)\) and     A1

at \(\left( {1,{\text{ }} - \sqrt {\frac{1}{2}} } \right)\) the tangent is \(y + \sqrt {\frac{1}{2}}  =  - \frac{{\sqrt 2 }}{2}(x - 1)\)     A1

Note: These equations simplify to \(y =  \pm \frac{{\sqrt 2 }}{2}x\).

Note: Award A0M1A1A0 if just the positive value of \(y\) is considered and just one tangent is found.

[4 marks]