Question 10:
10 | (a) | (𝑥 −2) = 5cosθ and (𝑦−1) = 5sinθ
(𝑥 −2)2 +(𝑦−1)2 = (5cos𝜃)2 +
(5sin𝜃)2 oe
(𝑥 −2)2 +(𝑦−1)2 = 52 oe isw
2 2
(𝑥−2) (𝑦−1)
or + = 1 oe isw
5² 5² | M1
M1
A1 | 3.1a
1.1
1.1 | allow sign errors
𝟐 𝟐
or ( 𝒙−𝟐 ) + ( 𝒚−𝟏 ) = cos2𝜃 +sin2𝜃 oe
𝟓 𝟓
may see eg 𝑥2 −4𝑥 +𝑦2 −2𝑦 = 20
if M0M0 allow SC1 for
𝑦 = 1+5sin (cos−1( 𝑥−2 ))
5
or 𝑥 = 2+5cos (sin−1( 𝑦−1 ))
5
[3]
Alternatively | if only seen in expanded form, allow one coefficient error;
allow sign errors
must have terms in cos𝜃 and sin𝜃
𝑥2 = (2+5cos𝜃)2and 𝑦2 = (1+5sin𝜃)2 | M1
𝑥2 +𝑦2 = 5+20cos𝜃 +10sin𝜃+ | M1
25sin2𝜃+25cos2𝜃
𝑥2 +𝑦2 = 20+4𝑥 +2𝑦 oe isw | A1
Alternatively
radius = 5 and centre is (2, 1) | M1 | allow sign error in coordinates of centre
(𝑥 −2)2 +(𝑦−1)2 = 52 | M1 | FT their centre
A1 | all correct
if only seen in expanded form, allow one coefficient error;
allow sign errors
must have terms in cos𝜃 and sin𝜃
10 | (b) | −4
gradient of radius is
3
3
gradient of tangent is
4
3
(𝑦−−3) = (𝑥 −5) oe
4
3𝑥 −4𝑦−27 = 0 𝑜𝑟−3𝑥 +4𝑦+27 = 0 | B1
M1
M1
A1 | 3.1a
2.1
2.4
1.1 | 4
FT 1÷𝑡ℎ𝑒𝑖𝑟−
3
3
allow one sign error; FT their
4
3
may see −3 = ×5+𝑐
4
[4]
Alternatively
d𝑦 5cos𝜃
= oe
d𝑥 −5sin𝜃 | B1
d𝑦 2−𝑥
or = oe
d𝑥 𝑦−1
d𝑦
eg 2(𝑥−2)+2(𝑦−1) = 0
d𝑥
3 4
substitution of cos𝜃 = and sin𝜃 = − oe
5 5 | M1 | 3⁄
d𝑦 = 5 𝐨𝐫 2−5 oe; allow one sign error;
d𝑥 −(−4⁄ ) −3−1
5 | 3⁄
d𝑦 = 5 𝐨𝐫 2−5 oe; allow one sign error;
d𝑥 −(−4⁄ ) −3−1
5
d𝑦
or (5,‒3) in their
d𝑥
3
(𝑦−−3) = (𝑥 −5) oe
4 | 3
(𝑦−−3) = (𝑥 −5) oe
4 | M1 | 3
allow one sign error; FT their
4
3
may see −3 = ×5+𝑐
4
3𝑥 −4𝑦−27 = 0 𝑜𝑟−3𝑥 +4𝑦+27 = 0 | A1
[4]
Alternatively
d𝑦 5cos𝜃
= oe
d𝑥 −5sin𝜃
B1