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)