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.