The curve C has equation \(2{x^2} + {y^2} = 18\). Determine the coordinates of the four points on C at which the normal passes through the point (1, 0) .

\(4x + 2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = -\frac{{2x}}{y}\)     M1A1

 Note: Allow follow through on incorrect \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) from this point.

gradient of normal at (a, b) is \(\frac{b}{{2a}}\)

 Note: No further A marks are available if a general point is not used

equation of normal at (a, b) is \(y - b = \frac{b}{{2a}}(x - a)\left( { \Rightarrow y = \frac{b}{{2a}}x + \frac{b}{2}} \right)\)     M1A1

substituting (1, 0)     M1    

\(b = 0\) or \(a = -1\)     A1A1

four points are \((3,{\text{ }}0),{\text{ }}( - 3,{\text{ 0}}),{\text{ }}( - 1,{\text{ }}4),{\text{ }}( - 1,{\text{ }} - 4)\)    A1A1 

 Note: Award A1A0 for any two points correct.

[9 marks]

Many students were able to obtain the first marks in this question by implicit differentiation but few were able to complete the question successfully. There were a number of students obtaining the correct final answers, but could not be given the marks due to incorrect working. Most common was students giving the equation of the normal as \(y - 0 = \frac{y}{{2x}}(x - 1)\), instead of taking a general point e.g. (a, b)