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 Further Mathematics Paper 2, Nov/Dec. 2013
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
Weakness/Remedies
Strength
Question 9

Question 9

Find the:
(a)        gradients of the tangents to the curve 2 + y2 – 2 – 6y + 5 = 0 at the points when  = 0.

(2)        equations of the tangents to the curve in 9(a).

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Observation

This question was reportedly attempted by majority of the candidates and they performed well in it.  Candidates were reported to have first obtained the y – coordinates of the points where x = 0 by substituting 0 for  in the given equation to obtain y2 – 6y + 5 = 0.  Solving this quadratic equation gave y = 1 and 5.  Differentiating 2 + y2 – 2 – 6y + 5 = 0 gave 2+ 2y - 2 - 6 = 0.  Bringing like terms together and simplifying gave  = .  Gradient to the curve at the point (0,1) was  at (0,1) =   -  =  The equation of the tangent to the curve x2 + y2 – 2x – 6y + 5 = 0 at (0,1) was y – 1 =   (x – 0) i.e. y – 1 = () which gave  + 2y – 2 = 0.  Similarly, the gradient of the curve at point (0,5) =  =  = .  Therefore, the equation of the tangent at (0,5)  = y- 5 = ½ () which gave  – 2y + 10 = 0.