Consider the curve with equation \({x^2} + xy + {y^2} = 3\).

(a)     Find in terms of k, the gradient of the curve at the point (−1, k).

(b)     Given that the tangent to the curve is parallel to the x-axis at this point, find the value of k.

(a)     Attempting implicit differentiation     M1

\(2x + y + x\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)     A1

EITHER

Substituting \(x = - 1,{\text{ }}y = k\,\,\,\,\,\)e.g. \( - 2 + k - \frac{{{\text{d}}y}}{{{\text{d}}x}} + 2k\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)     M1

Attempting to make \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) the subject     M1

OR

Attempting to make \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) the subject e.g. \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ - (2x + y)}}{{x + 2y}}\)     M1

Substituting \(x = - 1,{\text{ }}y = k{\text{ into }}\frac{{{\text{d}}y}}{{{\text{d}}x}}\)     M1

THEN

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2 - k}}{{2k - 1}}\)     A1     N1

(b)     Solving \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0{\text{ for }}k{\text{ gives }}k = 2\)     A1

[6 marks]

Part (a) was generally well answered, almost all candidates realising that implicit differentiation was involved. A few failed to differentiate the right hand side of the relationship. A surprising number of candidates made an error in part (b), even when they had scored full marks on the first part.