Question 10:
--- 10(a) ---
10(a) | 2
xa2 1 
 x63 20 or using x2y12
 
2  | *M1 | Obtaining an unsimplified equation in x or y only.
5
x2 32axa2 110
4 | A1 | OE e.g. 5x2 432ax4a2 44
Rearranging to get a correct 3-term quadratic on one side.
Condone terms not grouped together.
5y2  y544a133a2 24.
32a2 4 5 a2 11 0
4 | DM1 | OE Using b2 4ac on their 3 term quadratic 0.
Method 1 for final 2 marks
Using a = 4: 382 550 | A1 | Clearly substituting a = 4.
a16 | B1 | Condone no method shown for this value.
Method 2 for final 2 marks
a2 12a640 ⇒ a4a160 ⇒ a4 | A1 | AG Full method clearly shown.
a16 | B1 | Condone no method shown for this value.
5 | If M0, SCB1 available for substituting a4, finding
P(2, 7) and verifying that CP2 = 20.
Question | Answer | Marks | Guidance
--- 10(b) ---
10(b) | Centre (4, 3) identified or used or the point P is (2, 7) | B1
⸫ gradient of normal 2 | B1 | SOI
Forming normal equation using their gradient (not 0.5) and their centre or
P | M1 | Condone use of 4,3.
y3
2 or y72x2
x4 | A1 | OE Condone fx.
4
Question | Answer | Marks | Guidance
--- 10(c) ---
10(c) | Method 1 for Question 10(c)
1  1 
Diameter: y3 x4 leading to y x1
 
2  2 
Or
dy  1 
2x42y3 0 leading to y x1
 
dx  2  | *M1 | 1
Using gradient with their centre.
2
By implicit differentiation.
2
x42    1 x13   20   5 x2 10x0  
2  4  | DM1 | Obtaining an unsimplified equation in x or y only.
[y2 6y50].
x = 0 or 8, y = 1 or 5 [(0,1) and (8,5)] | A1 | Correct co-ordinates for both points. Condone no method
shown for solution.
Equations are y12x and y52x8 | A1 | 2x y1 and 2x y21.
Method 2 for Question 10(c)
Coordinates of points at which tangents meet curve are
(4+4,3+2) = (8,5) and (4 – 4,3 – 2) = (0,1) | *M1 A1 | Vector approach using their centre and gradient = 0.5 .
Condone answers only with no working.
Equations are y52x8 and y12x | DM1 A1 | Forming equations of tangents using their (0,1) and (8,5).
Method 3 for Question 10(c)
x42 2xc32
20
5x2 44cxc32 40
  | *M1 | Obtaining an unsimplified equation in x only using
equation of circle with y2xc.
44c2 20  c32 4  0
[leading to 4c2 32c120c161000] | DM1 | Using b2 4ac0.
Question | Answer | Marks | Guidance
10(c) | 4c2 88c840 [leading to c2 22c210 ] | A1
c21 and c1 or y2x21 and y2x1 | A1 | Condone no method shown for solution.
4
Question | Answer | Marks | Guidance