The distance between a point A(12,0) and B(0, -5) is found to be ………
a) 15 b) 18 c) 13 d) 25
AB = √((x2 – X1)2 + (y2- y1)2)
=√((0 -12)2 + (-5 -0))2
=√(144 + 25)
= √169 = 13 unit C
Find the direction of the vector V = 2i – 5j
a) 262o b) 272o c) 282o d) 292o
tanθ = (y/x)
tanθ = (-5/2)
θ = tan-1 (-5/2)
θ = -68.198
Since 2, -5 falls in the fourth quadrant:
θ = 360 – 68.198 = 292o D
A line which intercepts the y – axis at a point 1 and having a gradient of ¾ will have the following equation.
a) 3x – 4y = 4 b) 4x + 3y + 4 = 0 c) 3x – 4y = -4 d) 4x – 3y -4 = 0
y = mx + c
c =1 and m = ¾
y = ¾ x + 1
y = (3x + 4)/4
4y = 3x + 4
3x -4y = -4 C
Find the equation of a straight line joining the point M(-3, -4) and N(8, 1)
a) 5x + 11y + 29 =0 b) 5x – 11y -29 = 0 c) 11x + 5y = 29 d) 11x – 5y = -29
(y – y1)/(y2 – y1) = (x – x1)/(x2 – x1)
(y – (-4))/(1 – (-4)) = (x – (-3))/(8 – (-3))
(y + 4)/(1 + 4) = (x + 3)/(8 + 3)
( y + 4)/5 = (x + 3)/11
11(y+4) = 5(x + 3)
11y + 44 = 5x + 15
5x – 11y – 44 + 15 = 0
5x – 11y – 29 = 0 B
Write the equation of a straight line passing through the point (2, 2) and is parallel to the line 3x – 2y + 5 = 0
a)2y + 5x + 2 = 0 b) 3y – 2x = 2 c) 2y – 3x + 2 = 0 d) 3y + 2x -2 = 0
3x – 2y + 5 = 0
2y = 3x + 5
y =(3x + 5 )/2
y = 3x/2 + 5/2
Comparing with the formula Y = mx + c
m = 3/2 and c = 5/2
The slope of parallel line are equal to each other. Hence;
m1 = m2 = m = 3/2
Y – y1 = m (x – x1)
y – 2 = 3/2(x – 2)
2y – 4 = 3(x – 2)
2y – 4 = 3x – 6
2y -3x -4 + 6 = 0
2y -3x + 2 = 0 C
Solve V1 x V2 is V1 = -4i + 3j and V2 = i – 2j + 3k
Taking the cross product, we arrived at:
= i((3×3) –(1x – 2)) – j((-4 x 3) – (-1 x 1)) + k((-4 x 2) – (1 x 3))
=i( 9 + 2) –j(-12 -1) + k (8 – 3)
= i11 + 13j + 5k
Find the point of interception between the following line 4x – 9y + 9 = 0 and x + 14y – 79 = 0
a) (7, 3) b) (4, 2) c) ( 12, 5) d) (9, 5)
4x – 9y + 9 = 0 x 1
x + 14y – 79 = 0 x 4
4x – 9y + 9 = 0
4x + 56y -316 = 0
Subtracting the equation we arrive at
-65y + 325 = 0
65y = 325
y = 325/65
y = 5
Substituting the value of y in the equation above
4x – 9y + 9 = 0
4x – 9(5) + 9 = 0
4x – 45 + 9 = 0
4x = 36
x = 36/4 = 9
Hence (x, y) = (9, 5) D
Derive the equation of a circle whose center is at the origin and have a diameter of 10
a) x2 – y2 = 5 b) x2 + y2 = 25 c) 2x + 2y = 25 d) 2x – 2y = 5
d = 10 c(0, 0)
r = d/2 = 10/2 = 5
Using the equation of a circle: (x – α)2 + (y – β)2 = r2
(x – 0)2 + (y – 0)2 = 52
x2 + y2 = 25 B
Analyze the equation x2 + y2 -8x – 2y + 12 = 0 (Hint: Find the center and radius)
a) (5, 6) 6 b) (8, 4) 5 c) (4, 1) √2 d)(2, 2) 9
x2 + y2 -8x – 2y + 12 = 0
Using the equation of a circle x2 + y2 + 2gx + 2fy + c = 0. By comparison:
-8x = 2gx
g = -4
-2y = 2fy
f = -1
C = 12
C = g2 + f2 – r2
r2 = g2 + f2 – C
r2 = (-4)2 + (-1)2 -12
r2 = 16 + 1 -12
r2 = 5
r = √5
The center (a, b) = (-g, -f) = (-(-4), -(-1)) = (4, 1)
r = = √5
Hence: (4, 1) √5 C
Find (V1 x V2) . V3 V1 = 4i -8j + k, V2 = 2i +j – 2k, V3 = 3i – 4j + 12k
a) 235 b) 240 c)245 d) 250
Taking the cross product of V1 x V2 we have:
V1 x V2 = i(16 – 1) –j(-8 -2) + k(21 + 16) = 15i + j10 + 20k
Now taking the dot product (V1 x V2) . V3
= (15i + j10 + 20k) . (3i -4j + 12k)
=(45 – 40 + 240) = 245 C
Find the equation of a circle whose center is at (-3, 5/2) and having a radius of 3/2
a) x2 – y2 + 3x – 6y = 13 b) x2 + y2 + 5x + 6y = -13
c) x2 – y2 -3x + 6y = 13 d) x2 + y2 + 6x -5y = 13
C (-3, 5/2) and r 3/2
Using the equation of a circle x2 + y2 + 2gx + 2fy + c = 0
a = -g -g = -3
g = 3
b = -f -f = 5/2
f = -5/2
C = g2 + f2 – r2 = 32 + (-5/2)2 – (3/2)2
C = 9 + 25/4 – 9/4 = 52/4 = 13
Now inserting the value of g, f and c in the general equation:
x2 + y2 + 2gx + 2fy + c = 0
x2 + y2 + 2(3)x + 2(-5/2)y + 13 = 0
x2 + y2 + 6x -5y + 13 = 0 C
Find the focus and directrix of the following parabola x2 = 28y
a) (9, 4) x=9 b)(7, 0), x= 7 c) (28, 14), x= 28 d) (14, 0), x=14
x2 = 28y
Comparing this to the general equation of parabola x2 = 4ay
4a = 28
a = 28/4 =7
Hence F(7, 0) x=7 B
Derive the equation of an ellipse whose focus is at F(5, 0) and having major axis which is a=8.
a) x2/100 + y2/36 = 1 b) x2/100 – y2/36 = 1 c) x2/64 + y2/36 = 1 d) x2/64 – y2/36 = 1
F(5, 0) and major axis a = 8
x = ae =5
e = 5/a = 5/8
b2 = a2 (1 – e2)
b2 = 82 –( 1- (5/8)2)
b2 = 64(1 – 25/64)
b2 = 39
Using the equation of parabola at origin x2/a2 + y2/b2 = 1
x2/82 + y2/39 = 1
x2/64 + y2/39 = 1 C
A hyperbolic equation is given as x2/16 – y2/49 = 1. Its transvers and coordinate systems are:
a) 4 & 49 b) 16 & 49 c) 16 & 7 d) 8 & 14
x2/16 – y2/49 = 1
length of transvers = 2a
length of conjugate axis = 2b
√a = √16 = 4
√b = √49 = 7
L.T = 2a = 2 x 4 = 8
L.C = 2b = 2 x 7 = 14
Analyse the following parabola y2 – 4y + 12x + 16 = 0
a) V(-4, 12), F(-4, 8) b) V(2, 1), F(8, 1) c) V(-1, 2), F(-4, 2) d) V(6, 3), F(6, 6)
y2 – 4y + 12x + 16 = 0
Solving parabola y2 – 4y using complex square
y2 – 4y = (y -2)2 -4
(y -2)2 -4 + 12x + 16 = 0
(y -2)2 -4 = -12x – 16
(y -2)2 = -12x – 16 + 4
(y -2)2 = -12x – 12
(y -2)2 = -12(x + 1)
by comparing with the formula (y -k)2 = 4a(x – h)
4a = -12 a = -3
-k = -2 k = 2
-h = 1 h = -1
(h, K) = (-1, 2) = Vertex
Focus = F(a+h, k) = (-3 -1, 2) = F(-4, 2) C
Find the slope of the line joining (-9, -6) and (0, 2)
a) 9/5 b) -3 c) -8/9 d) 3/2
Slope = (y2 – y1)/ (X2 – X1)
= (2 – -6)/0 – -9) = 2+6/ 0 + 9
=8/9 C
Find the angle made by the vector V = 2i + 4j with the axis
a) 63o b) 64o c) none d) 65o
tanθ = y/x = 4/2
θ= tan-1 (4/2)
θ = 63o A
Find the angle made by the vector V = 2i + 4j with the axis
a) 63o b) 64o c) none d) 65o
tanθ = y/x = 4/2
θ= tan-1 (4/2)
θ = 63o A
- 1a) Find the centre, vertices, foci, eccentricity and asymptotes of the hyperbola with equation
9x2 – 4y2 – 18x + 32y -91 = 0
- 1b. The parametric equations of a circle are x = 5cost and y=5sint. Find its Cartesian equation
- 1c.Write an equation of the circle shown in the figure below.
- a)
9𝑥2−4𝑦2−18𝑥+32𝑦−91=0
(9𝑥2−18𝑥)−(4𝑦2−32𝑦)−91=0
9(𝑥2−2𝑥)−4(𝑦2−8𝑦)−91=0
Adding square of the half coefficient of x and y
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥=(2/2)2=1
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑦=(8/2)2=16
9(𝑥2−2𝑥+1−1)−4(𝑦2−8𝑦+16−16)−91=0
9(𝑥2−2𝑥+1)−9−4(𝑦2−8𝑦+16)+64−91=0
9(𝑥2−𝑥−𝑥+1)−4(𝑦2−4𝑦−4𝑦+16)−9+64−91=0
9(𝑥(𝑥−1)−1(𝑥−1))−4(𝑦(𝑦−4)−4(𝑦−4))−36=0
9(𝑥−1)2−4(𝑦−4)2=36
9/36(𝑥−1)2 − 4/36(𝑦−4)2=36/36
1/4(𝑥−1)2−1/9(𝑦−4)2=1
(𝑥−ℎ)2 /𝑎2 − (𝑦−𝑘)2 /𝑏2=1
𝐶𝑒𝑛𝑡𝑒𝑟 (ℎ,𝑘)=(1,4) 𝑎2=4 𝑎=2 𝑏2=9 𝑏=3
𝑉𝑒𝑟𝑡𝑖𝑐𝑒𝑠 (ℎ±𝑎, 𝑘)=(1±2, 4)
𝑐2=𝑎2+𝑏2 𝑐=√(4+9)= 𝑐=√13
𝐹𝑜𝑐𝑖𝑠(ℎ±𝑐, 0)= (1±√13, 0)
𝑒=[√(𝑎2+𝑏2) ]/ 𝑎
e=[√(4+9)]/2=√13/2
𝐴𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑠 𝑦 = ±(𝑏/𝑎)(𝑥−ℎ)+𝑘 = ±(3/2)(𝑥−1) + 4
- b)
𝑥=5cos𝑡 𝑦=5sin𝑡
sin2𝜃+cos2𝜃=1
𝑥2=25cos2𝑡 cos2𝑡=𝑥2 /25 ………….(𝑖)
𝑦2=25sin2𝑡 sin2𝑡=𝑦2/25 ………….(𝑖𝑖)
sin2𝑡+cos2𝑡=1…….(𝑖𝑖𝑖)
Sub eqn (i) and (ii) in (iii)
𝑦2 /25+𝑥2 /25=1
𝑦2+𝑥2=25
- c) The centre of the circle is at (−2,2)
The radius is 3
(𝑥−ℎ)2+(𝑦−𝑘)2=𝑟2
(𝑥+2)2+(𝑦−2)2=32
(𝑥+2)2+(𝑦−2)2=9
- a) If 𝑢̅=(2𝑡2𝑖+3𝑗) 𝑎𝑛𝑑 𝑣̅=(𝑡3𝑖+7𝑡𝑗) verify the result
- 𝑑/𝑑𝑡(𝑢̅.𝑣̅)=𝑢̅.𝑑𝑣̅/𝑑𝑡+𝑑𝑢̅/𝑑𝑡.𝑣̅
- 𝑑/𝑑𝑡(𝑢̅ × 𝑣̅)= 𝑢̅ × 𝑑𝑣̅/𝑑𝑡+𝑑𝑢̅/𝑑𝑡×𝑣̅
- b) Find the coordinates of the centre, vertices and foci of the ellipse
25x2 + y2 = 25
tanθ = y/x = 4/2
θ= tan-1 (4/2)
θ = 63o A
- a)
i) 𝑑𝑣̅/𝑑𝑡=3𝑡2𝑖+7𝑗 𝑑𝑢̅/𝑑𝑡=4𝑡𝑗
𝑢̅.𝑣̅ = (2𝑡2𝑖+3𝑗).(𝑡3𝑖+7𝑡𝑗) = 2𝑡5+21𝑡
𝑑/𝑑𝑡(𝑢̅.𝑣̅) =10𝑡4+21
𝑢̅.𝑑𝑣̅ / 𝑑𝑡= (2𝑡2𝑖+3𝑗).(3𝑡2𝑖+7𝑗)= 6𝑡4+21
𝑑𝑢̅/𝑑𝑡 . 𝑣̅ = (4𝑡𝑗).(𝑡3𝑖+7𝑡𝑗)=4𝑡4
𝑢̅.𝑑𝑣̅/𝑑𝑡+𝑑𝑢̅/𝑑𝑡.𝑣̅ = 6𝑡4+21+4𝑡4 = 10𝑡4+21
𝑇ℎ𝑢𝑠 𝑑/𝑑𝑡(𝑢̅.𝑣̅) = 𝑢̅.𝑑𝑣̅/𝑑𝑡+𝑑𝑢̅/𝑑𝑡.𝑣
- ii) 𝑑𝑣̅ /𝑑𝑡 = 3𝑡2𝑖 + 7𝑗 𝑑𝑢̅ /𝑑𝑡 = 4𝑡𝑗
i j k
𝑢̅ × 𝑣̅ = 2t2 3 0
t3 7t 0
= 𝑜𝑖 + 0𝑗 + (14𝑡3 − 3𝑡3)𝑘 = 11𝑡3𝑘
𝑑 /𝑑𝑡 (𝑢̅ × 𝑣̅) = 33𝑡2𝑘
i j k
𝑢̅ × 𝑑𝑣̅ /𝑑𝑡 = 2t2 3 0
t3 7 0
= 0𝑖 + 0𝑗 + (14𝑡2 − 9𝑡2)𝑘 = 5𝑡2𝑘
i j k
𝑑𝑢̅ /𝑑𝑡 × 𝑣̅= 2t2 0 0
t3 7 0
= 0𝑖 + 0𝑗 + 28𝑡2𝑘 = 28𝑡2𝑘
𝑢̅ × 𝑑𝑣̅ /𝑑𝑡 + 𝑑𝑢̅ /𝑑𝑡 × 𝑣̅ = 5𝑡2𝑘 + 28𝑡2𝑘 = 33𝑡2𝑘
𝑇ℎ𝑢𝑠 𝑑/ 𝑑𝑡 (𝑢̅ × 𝑣̅) = 𝑢̅ × 𝑑𝑣̅ /𝑑𝑡 + 𝑑𝑢̅ /𝑑𝑡 × 𝑣̅
- b) 25𝑥2 + 𝑦2 = 25
Divide the equation by 25
25𝑥2 /25 + 𝑦2 /25 = 25 /25
𝑥2 /1 + 𝑦2 /25 = 1
𝑥2/ 𝑏2 + 𝑦2/ 𝑎2 = 1
𝑎 > 𝑏 𝑐2 = 𝑎2 − 𝑏2 𝑎2 = 25 𝑎 = 5
𝑏2 = 1 𝑏 = 1
𝑐 = √(𝑎2 − 𝑏2) = √(25 – 1) = √24
= √(4 × 6) = 2√6
𝐶𝑒𝑛𝑡𝑒𝑟 𝑎𝑡 𝑜𝑟𝑖𝑔𝑖𝑛 (0,0)
𝑉𝑒𝑟𝑡𝑖𝑐𝑒𝑠 (0, ±𝑎) = (0, ±5)
𝐹0𝑐𝑖 (0, ±𝑐) = (0, ±2√6)
- a) Let a and b be two-dimensional vector is the following result true?
Where x denotes the vector product
- b) Find the equation of the hyperbola with vertices at (0,+-6) and e=5/3. Find the foci
- c. What is the point of intersection of the lines with equation
x=7 and y =-9?
- b)
𝑒=√(𝑎2+𝑏2 ) 𝑎=√[(𝑎2+𝑏2 )/𝑎2 ]
𝑦2 /𝑎2 − 𝑥2 /𝑏2=1 𝑣(0,±𝑎)
𝑥2 /𝑎 2− 𝑦2 /𝑏2=1 𝑣(±𝑎,0)
𝑎=6 𝑒=53
5/3= √[(62+𝑏2 )/62]
5/3= √[(36 + 𝑏2 )/36]
25/9=(36+𝑏2 )/36
900=324+9𝑏2
9𝑏2=900−324
𝑏2=64 𝑏 = √64 = 8
𝑇ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑦𝑝𝑒𝑟𝑏𝑜𝑙𝑎 𝑦2 /62 − 𝑥2/82=1
𝑦2/36 − 𝑥2/64=1 𝑐2=𝑎2+𝑏2 𝑐=√36+64 𝑐=√100=10
𝐹𝑜𝑐𝑖(0,±𝑐) = (0,±10)
- c) 𝑥 = 7 𝑎𝑛𝑑 𝑦 =−9
𝑇ℎ𝑢𝑠 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡 (7, −9)
The equation of a circle is x2 + y2 – x – 3y = 0. The line y = x – 1 Intersect the circle at point P and Q
- i) Find the coordinates of P and Q
- ii) Sketch and show the points P and Q
- iii) Determine the length of PQ
- vi) Determine the coordinate of M, the mid-point of line PQ
- v) If O is the centre of the circle, show that PQ is perpendicular to OM
- vi) Determine the equation of the tangent to the circle at P and Q
- vii) Find the point where the two tangent intersect
- i) 𝑥2+𝑦2−𝑥−3𝑦=0………….(𝑖)
𝑦=𝑥−1…………………………(𝑖𝑖)
Substitute equation (ii) in (I)
𝑥2+(𝑥−1)2−𝑥−3(𝑥−1)=0
𝑥2+𝑥2−2𝑥+1−𝑥−3𝑥+3=0
2𝑥2+−6𝑥+4=0
𝑥2+−3𝑥+2=0
𝑥2+−2𝑥−𝑥+2=0
𝑥(𝑥−2)−1(𝑥−2)=0
(𝑥−2)(𝑥−1)=0
𝑥=2 𝑎𝑛𝑑 𝑥=1
𝑦=𝑥−1 𝑦=2−1=1
𝑦=1−1 =0
𝑦=1 𝑎𝑛𝑑 𝑦=0
The two points are 𝑃(1,0) 𝑄(2,1)
- ii)
- iii) Distance
𝑑=√[(𝑥2−𝑥1)2+(𝑦2−𝑦1)2
√[(2−1)2+(1−0)2]
d=√1+1=√2= √2
- iv) Mid-point M
𝑀𝑦=(𝑦2+𝑦1)/2
𝑀𝑥=(𝑥2+𝑥1)/2
𝑀𝑦=(1+0)/2=1/2
𝑀𝑥=(2+1)/2=3/2
𝑀(3/2, 1/2)
- v)𝑃𝑄 𝑃𝑒𝑟𝑝𝑒𝑚𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑂𝑀 𝑤ℎ𝑒𝑟𝑒 𝑂 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒
𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑙𝑖𝑛𝑒𝑠=−1 𝑚1𝑚2=−1
The centre of the circle from the equation 𝑂(1/2, 3/2)
𝑃(1, 0), 𝑄(2, 1), 𝑂(1/2, 3/2) 𝑎𝑛𝑑 𝑀(3/2, 1/2)
𝑚𝑃𝑄 𝑚𝑂𝑀=−1
𝑚𝑃𝑄=(𝑦2−𝑦1 )/(𝑥2−𝑥1 )
𝑚𝑃𝑄=(1−0)/(2−1)=1
𝑚𝑂𝑀=[(1/2−3/2)/ (3/2−1/2)]=−1
𝑚𝑃𝑄 𝑚𝑂𝑀= 1×−1 =−1 𝑇ℎ𝑢𝑠 𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟
- vi) Equation of P and Q
𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑡ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 𝑂𝑃
𝑂(1/2, 3/2) 𝑃(1,0)
𝑚𝑂𝑝=(1−1/2)/ (0−3/2)=−3
The Tangent is perpendicular to the radius
𝑇ℎ𝑢𝑠 −3×𝑚𝑝=−1
𝑚𝑝=1/3 𝑃(1,0)
(𝑦−𝑦1)=𝑚(𝑥−𝑥1)
𝑦−0=1/3(𝑥−1)
𝑦=𝑥/3−1/3
𝑂(1/2, 3/2) 𝑄(2,1)
𝑚𝑂𝑄=(1−3/2)/ (2−1/2) =1/−3
The Tangent is perpendicular to the radius
𝑇ℎ𝑢𝑠 1−3×𝑚𝑄=−1
𝑚𝑄=3 𝑄(2,1)
(𝑦−𝑦1)=𝑚(𝑥−𝑥1)
𝑦−1=3(𝑥−2) 𝑦−1=3𝑥−6
𝑦 = 3𝑥−5
- vii) The point of intercept
𝑦 = 𝑥/3 – 1/3………….(𝑖)
𝑦 = 3𝑥 − 5……….(𝑖𝑖)
𝑦=𝑦
𝑥/3 – 1/3 = 3𝑥−5
𝑥 – 1 = 9𝑥−15
9𝑥 – 𝑥 = 15−1
14=8𝑥
𝑥=14/8 = 7/4
The points of intercept are (7/4, 1/4)
- a) Find the equation of the circle passing through three points (1, 6) (2, 1) and (5, 2). Also, find the coordinates of its centre and the length of the radius
- b) Prove that the circle x2 + y2 – 16y + 32 = 0 and x2 + y2 +18x + 2y + 32 = 0 touch externally
- c) Calculate the area of the parallelogram spanned by the vectors a and b where a = (1,3,-3) and b =(8,18,4)
tanθ = y/x = 4/2
θ= tan-1 (4/2)
θ = 63o A
- a)
(1, −6) (2, 1) 𝑎𝑛𝑑 (5, 2)
𝑥2 + 𝑦2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐=0
𝐹𝑜𝑟 (1,−6)
12 + (−6)2 + 2𝑔 + 2𝑓 × −6 + 𝑐 = 0
1 + 36 + 2𝑔 − 12𝑓 + 𝑐 = 0
2𝑔 − 12𝑓 + 𝑐 = −37 … … … . . (𝑖)
𝐹𝑜𝑟 (2,1)
22 + 12 + 2 × 2𝑔 + 2𝑓 + 𝑐 = 0
5 + 1 + 2𝑔 + 2𝑓 + 𝑐 = 0
2𝑔 + 2𝑓 + 𝑐 = −5 … … … … . . . (𝑖𝑖)
𝐹𝑜𝑟 (5, 2)
52 + 22 + 5 × 2𝑔 + 2 × 2𝑓 + 𝑐 = 0
25 + 4 + 10𝑔 + 4𝑓 + 𝑐 = 0
10𝑔 + 4𝑓 + 𝑐 = −29 … … … … . . . (𝑖𝑖𝑖)
Subtracting eqn (i) from (ii)
2𝑔 + 14𝑓 = 32 … … … . . (𝑖𝑣)
Subtracting eqn (i) from (iii)
8𝑔 + 16𝑓 = 8 … … … … (𝑣)
𝑒𝑞𝑛(𝑖𝑣) 𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 2 𝑔 + 7𝑓 = 16 … . . (𝑣𝑖)
𝑒𝑞𝑛(𝑣) 𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 8 𝑔 + 2𝑓 = 1 … . . (𝑣𝑖)
Subtracting g + 2f = 1 from g + 7f = 16
We arrive at 5f = 15
f = 3
𝑔 = 1 − 2𝑓 = 1 − 2(3) = −5
𝑔 = −5
𝐹𝑟𝑜𝑚 𝑒𝑞𝑛 (𝑖𝑖)
4𝑔 + 2𝑓 + 𝑐 = 5
𝑐 = −5 − 4𝑔 − 2𝑓 = −5 − 4(−5) − 2(3)
𝑐 = −5 + 20 − 6 = 9
𝑔 = −5 𝑓 = 3 𝑐 = 9
𝑥2 + 𝑦2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0
𝑥2 + 𝑦2 + 2(−5)𝑥 + 2(3)𝑦 + 9 = 0
𝑥2 + 𝑦2 − 10𝑥 + 6𝑦 + 9 = 0
𝑇ℎ𝑒 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑒𝑟 (−𝑔, −𝑓) = (5, −3)
𝑅𝑎𝑑𝑖𝑢𝑠 = √(𝑔2 + 𝑓2)− 𝑐 = √(25 + 9 – 9) = √25 = 5 𝑈𝑛𝑖𝑡𝑠
- b) 𝑥2 + 𝑦2 − 16𝑦 + 32 = 0 and 𝑥2 + 𝑦2 − 18𝑥 + 2𝑦 + 32 = 0
𝑥2 + 𝑦2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0
1𝑠𝑡 𝐶𝑖𝑟𝑐𝑙𝑒
𝑥2 + 𝑦2 − 16𝑦 + 32 = 0
𝑔 = 0 𝑓 = −8
𝑇ℎ𝑒 𝑐𝑜𝑜𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑒𝑟 (−𝑔, −𝑓) = (0, 8)
𝑟1 = √(02 + 82 – 32) = √(64 – 32) = √32 = 5.66
2𝑛𝑑 𝐶𝑖𝑟𝑐𝑙𝑒
𝑥2 + 𝑦2 − 18𝑥 + 2𝑦 + 32 = 0
𝑔 = −9 𝑓 = 1
𝑇ℎ𝑒 𝑐𝑜𝑜𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑒𝑟 (−𝑔, −𝑓) = (9, 1)
𝑟2 = √((−9)2 + 12 – 32) == √50 = 7.071
𝑑 = 𝑟1 + 𝑟2 = 5.66 = 7.071 = 12.731
𝑑 = √{(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2 }
𝑑 = √[(9 − 0)2 + (−1 − 8)2] = √(92 + 92) = √162 = 12.727
𝑆𝑖𝑛𝑐𝑒 𝑟1 + 𝑟2 = √{(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2 } 𝑇ℎ𝑢𝑠, 𝑡ℎ𝑒𝑦 𝑡ℎ𝑜𝑢𝑐ℎ 𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙𝑙𝑦
- c)
tanθ = y/x = 4/2
θ= tan-1 (4/2)
θ = 63o A
- a) A circle passes through point A(2,4) and B(-2,6) and the center lies on the line x+ 3y -8 = 0
- b) A line is draw perpendicular to 4x – 3y + 1 = 0 through the point (1,3) and (1,3)
- a) The intersect of the chord AB bisector Z and the given line is the centre M of the circle and the bisector is normal through the mid-point
My = (y2 + y1)/2 = (6+4)/2 = 10/2 =5
Mx = (x2 + x1)/2 = (2-2)/2 = 0/2 =0
M(0,5)
The bisector is perpendicular to the line AB
mABmZ = -1
mAB = (6-4)/(-2-2) = 2/(-4) = -1/2
mAB = -1/mZ =-1/(-1/2) = 2
Equation of the bisector (0,5)
(y – y1 ) = m(x – x1)
(y – 5) = 2(x-0)
y = 2x + 5
y – 2x = 5…….×3
3y – 6x = 15……….(I)
From the question equation of the line is x + 3y – 8 = 0
3y + x = 8…………..(ii)
Subtracting equation (ii) from (I)
We will arrive at
-7x = 7 x = -1
Substituting the value of x in y -2x = 5
y = 2(-1) + 5 = -2 + 5 = 3
Z(-1, 3)
The radius of the line is the distance between AZ or BZ
r=d(BZ)=√((-1+2)2 + (3-6)2 ) = √(1+9) = √10
Formula of a circle (x-p)2 +(y-q)2 =r2
(x+1)2 + (y-3)2 = (√10)2
(x+1)2 + (y-3)2 = 10
- b) 4𝑥−3𝑦+1=0
(1,3) and (6,𝑘)
4𝑥+1=3𝑦 𝑦=4𝑥/3 + 13
𝑦=𝑚𝑥+𝑐
𝑚1=4/3
𝑚1𝑚2=−1
𝑚2=−1/(4/3) =−3/4
𝑚2=(𝑦2−𝑦1 )/(𝑥2−𝑥1 )
−3/4 = (𝑘−3)/(6−1)
−3/4 = (𝑘−3)/5
−15=4(𝑘−3)
−15=4𝑘−12 4𝑘=−3
𝑘=−3/4
tanθ = y/x = 4/2
θ= tan-1 (4/2)
θ = 63o A
- a) Write an equation for the ellipse having one focus at (0,3) at a vertex (0,4) and it’s centre at (0,0)
- b) What is eccentricity and what are the ranges of values of eccentricity for circle, ellipse, parabola and hyperbola?
- c) A particle is moving on the x-axis. At time t = 0 the particle is at the point where x = 5 The velocity of the particle at time t seconds (where t>= 0) is (6t2 – t2) ms-1 Find:
- i) The value of t for which P is instantaneously
- ii) The acceleration of the particle when t=2 and when t=4
- iii) An expression for the displacement of the particle from O at time t second s
- a) Ellipse
𝐹𝑜𝑐𝑢𝑠(0,3) 𝑉𝑒𝑟𝑡𝑒𝑥(0,4) 𝐶𝑒𝑛𝑡𝑒𝑟(0,0)
𝑦2/𝑎2 + 𝑥2/𝑏2=1 𝑣(0,±𝑎)
𝑎=4 𝑐=3 𝑐2=𝑎2−𝑏2
𝑏2=𝑎2−𝑐2=√(42−32)=√(16−9)=√7
𝑦2/42 + 𝑥2/(√7)2=1
𝑦2/16 + 𝑥2/7=1
- b) Eccentricity: how much a conic section (a circle, ellipse, parabola or hyperbola) varies from being circular.
- At e = 0 for Circle
- At 0 < e < 1 for Ellipse
- At e = 1 for Parabola
- At e > 1 for hyperbola
- At e = Infinity for a line
- c) If the particle is at rest
i) v(t) = 0
v(t) = 6t – t2 = 0
t(6 –t) = 0
6 – t = 0 t = 6 sec
ii) a(t) = dv/dt = d(6t – t2)/dt
a(t) = 6 – 2t
when t = 2 a(t) = 6 -2(2) = 6 – 4 = 2 m/s2
iii) x(t) = ∫ v dt = ∫(6t – t2)dt
x(t) = 3t2 – t3/3 + c
when t=0 and x(t) = 6
6=3(0) -0/3 + c
c =6
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