Exercise 10.1 — Circles
Circle with centre (0,2) and radius 2.
Answer
(x–0)² + (y–2)² = 4
⇒ x² + (y–2)² = 4
⇒ x² + y² – 4y = 0
Circle with centre (–2,3) and radius 4.
Answer
(x+2)² + (y–3)² = 16
⇒ x² + y² + 4x – 6y – 3 = 0
Circle with centre (1/2, 1/4) and radius 1/12.
Answer
(x–1/2)² + (y–1/4)² = 1/144
⇒ x² + y² – x – y/2 + 1/4 + 1/16 – 1/144 = 0
Standard form: (x–1/2)² + (y–1/4)² = 1/144
Circle with centre (1,1) and radius √2.
Answer
(x–1)² + (y–1)² = 2
⇒ x² + y² – 2x – 2y = 0
Circle with centre (–a,–b) and radius √(a²–b²).
Answer
(x+a)² + (y+b)² = a²–b²
→ x² + y² + 2ax + 2by + 2b² = 0
Find centre and radius of (x+5)² + (y–3)² = 36.
Answer
Centre = (–5, 3), radius = 6
Find centre and radius of x² + y² – 4x – 8y – 45 = 0.
Answer
(x²–4x+4) + (y²–8y+16) = 45+4+16 = 65
(x–2)² + (y–4)² = 65
⇒ Centre = (2, 4), radius = √65
Find centre and radius of x² + y² – 8x + 10y – 12 = 0.
Answer
(x–4)² + (y+5)² = 12+16+25 = 53
⇒ Centre = (4,–5), radius = √53
Find centre and radius of 2x² + 2y² – x = 0.
Answer
x² + y² – x/2 = 0
⇒ (x–1/4)² + y² = 1/16
Centre = (1/4, 0), radius = 1/4
Circle passing through (4,1) and (6,5) with centre on 4x + y = 16.
Answer
Let centre = (h,k). Since equidistant from (4,1) and (6,5):
(h–4)²+(k–1)² = (h–6)²+(k–5)²
h²–8h+16+k²–2k+1 = h²–12h+36+k²–10k+25
4h+8k = 44 → h+2k = 11 ...(1)
Also 4h+k = 16 ...(2)
From (1) and (2): 4h+k=16 and h+2k=11
⇒ h=3, k=4
r² = (3–4)²+(4–1)² = 1+9 = 10
Equation: (x–3)² + (y–4)² = 10
⇒ x² + y² – 6x – 8y + 15 = 0
Circle passing through (2,3) and (–1,1) with centre on x–3y–11 = 0.
Answer
Equidistant from points: (h–2)²+(k–3)² = (h+1)²+(k–1)²
⇒ –4h+4–6k+9 = 2h+1–2k+1
⇒ –6h–4k+12 = 0
⇒ 3h+2k = 6 ...(1)
Centre on x–3y–11=0: h–3k = 11 ...(2)
From (1) and (2): h = 7, k = –7/2. Wait: let me solve again.
3h+2k=6 and h–3k=11
⇒ multiply (2) by 2: 2h–6k=22. Multiply (1) by 3: 9h+6k=18.
Adding: 11h = 40 → h = 40/11...
Actually: from h = 11+3k; substitute: 3(11+3k)+2k=6
⇒ 33+11k=6
⇒ k=–3; h=2
r² = (2–2)²+(–3–3)² = 36. Equation: (x–2)² + (y+3)² = 36
⇒ x²+y²–4x+6y–23=0
Circle with radius 5, centre on x-axis, passing through (2,3).
Answer
Centre = (h,0). (h–2)²+9 = 25
⇒ (h–2)² = 16
⇒ h=6 or h=–2
Equations: (x–6)²+y²=25 or (x+2)²+y²=25
Circle passing through (0,0) making intercepts a and b on axes.
Answer
Circle passes through (0,0), (a,0), (0,b). General equation x²+y²+Dx+Ey+F=0.
F=0; D=–a; E=–b. Equation: x²+y²–ax–by=0
Circle with centre (2,2) passing through (4,5).
Answer
r² = (4–2)²+(5–2)² = 4+9 = 13. Equation: (x–2)²+(y–2)²=13
Does (–2.5, 3.5) lie inside, outside, or on x²+y²=25?
Answer
(–2.5)²+(3.5)² = 6.25+12.25 = 18.5 < 25. Point lies INSIDE the circle.
Exercise 10.2 — Parabola
y² = 12x: focus, axis, directrix, latus rectum.
Answer
4a=12
⇒ a=3. Focus: (3,0). Axis: x-axis. Directrix: x=–3. LR = 12.
x² = 6y: focus, axis, directrix, latus rectum.
Answer
4a=6
⇒ a=3/2. Focus: (0,3/2). Axis: y-axis. Directrix: y=–3/2. LR=6.
y² = –8x: focus, axis, directrix, latus rectum.
Answer
4a=8
⇒ a=2. Opens left. Focus: (–2,0). Axis: x-axis. Directrix: x=2. LR=8.
x² = –16y: focus, axis, directrix, latus rectum.
Answer
4a=16
⇒ a=4. Opens down. Focus: (0,–4). Axis: y-axis. Directrix: y=4. LR=16.
y² = 10x: focus, axis, directrix, latus rectum.
Answer
4a=10
⇒ a=5/2. Focus: (5/2,0). Axis: x-axis. Directrix: x=–5/2. LR=10.
x² = –9y: focus, axis, directrix, latus rectum.
Answer
4a=9
⇒ a=9/4. Opens down. Focus: (0,–9/4). Axis: y-axis. Directrix: y=9/4. LR=9.
Equation of parabola: Focus (6,0), directrix x=–6.
Answer
a=6 (focus on x-axis, right). y²=4(6)x = 24x
Focus (0,–3), directrix y=3.
Answer
a=3, opens down. x²=–4(3)y = –12y
Vertex (0,0), focus (3,0).
Answer
a=3, axis x-axis. y²=12x
Vertex (0,0), focus (–2,0).
Answer
a=2, opens left. y²=–8x
Vertex (0,0), axis along x-axis, passing through (2,3).
Answer
y²=4ax. 9=8a
⇒ a=9/8. y²=4(9/8)x = 9x/2
⇒ 2y²=9x
Vertex (0,0), symmetric about y-axis, passing through (5,2).
Answer
x²=4ay. 25=8a
⇒ a=25/8. x²=4(25/8)y = 25y/2
⇒ 2x²=25y
Exercise 10.3 — Ellipse
x²/36 + y²/16 = 1.
Answer
a²=36, b²=16, c²=a²–b²=20, c=2√5. Foci: (±2√5, 0). Vertices: (±6, 0). Eccentricity: e=2√5/6=√5/3. LR = 2b²/a = 32/6 = 16/3.
x²/4 + y²/25 = 1.
Answer
a²=25 (major on y-axis), b²=4, c²=21, c=√21. Foci: (0,±√21). Vertices: (0,±5). e=√21/5. LR = 2b²/a = 8/5.
x²/16 + y²/9 = 1.
Answer
a²=16, b²=9, c²=7, c=√7. Foci: (±√7, 0). Vertices: (±4,0). e=√7/4. LR=2(9)/4=9/2.
x²/25 + y²/100 = 1.
Answer
a²=100 (major on y), b²=25, c²=75, c=5√3. Foci: (0,±5√3). Vertices: (0,±10). e=5√3/10=√3/2. LR=2(25)/10=5.
x²/49 + y²/36 = 1.
Answer
a²=49, b²=36, c²=13. Foci: (±√13, 0). Vertices: (±7,0). e=√13/7. LR=2(36)/7=72/7.
x²/100 + y²/400 = 1.
Answer
a²=400 (y-axis), b²=100, c²=300, c=10√3. Foci: (0,±10√3). Vertices: (0,±20). e=√3/2. LR=2(100)/20=10.
36x² + 4y² = 144.
Answer
Divide by 144: x²/4 + y²/36 = 1. a²=36, b²=4, c²=32, c=4√2. Foci: (0,±4√2). LR=2(4)/6=4/3.
16x² + y² = 16.
Answer
x² + y²/16 = 1. a²=16, b²=1, c²=15, c=√15. Foci: (0,±√15). LR=2(1)/4=1/2.
4x² + 9y² = 36.
Answer
x²/9 + y²/4 = 1. a²=9, b²=4, c²=5, c=√5. Foci: (±√5,0). LR=2(4)/3=8/3.
Vertices (±5,0), foci (±4,0).
Answer
a=5, c=4, b²=a²–c²=9. Equation: x²/25 + y²/9 = 1
Vertices (0,±13), foci (0,±5).
Answer
a=13, c=5, b²=144. x²/144 + y²/169 = 1
Vertices (±6,0), foci (±4,0).
Answer
a=6, c=4, b²=20. x²/36 + y²/20 = 1
Ends of major axis (±3,0), ends of minor axis (0,±2).
Answer
a=3, b=2. x²/9 + y²/4 = 1
Ends of major axis (0,±√5), ends of minor axis (±1,0).
Answer
a=√5, b=1. x² + y²/5 = 1
Length of major axis 26, foci (±5,0).
Answer
a=13, c=5, b²=144. x²/169 + y²/144 = 1
Length of minor axis 16, foci (0,±6).
Answer
b=8, c=6, a²=64+36=100. x²/64 + y²/100 = 1
Foci (±3,0), a=4.
Answer
c=3, a=4, b²=7. x²/16 + y²/7 = 1
b=3, c=4, major axis along x-axis.
Answer
a²=b²+c²=9+16=25. x²/25 + y²/9 = 1
Centre at origin, major axis along y-axis, a=6, passes through (4,–2).
Answer
x²/b² + y²/36 = 1. 16/b² + 4/36 = 1
⇒ 16/b² = 8/9
⇒ b²=18. x²/18 + y²/36 = 1
Major axis along x-axis, passes through (4,3) and (–1,4).
Answer
x²/a² + y²/b² = 1. 16/a² + 9/b² = 1 and 1/a² + 16/b² = 1
Let u=1/a², v=1/b²: 16u+9v=1 and u+16v=1
16(u+16v)–(16u+9v)=16–1
⇒ 247v=15
⇒ v=15/247
u=1–16(15/247)=(247–240)/247=7/247
a²=247/7, b²=247/15. Equation: 7x²/247 + 15y²/247 = 1
Exercise 10.4 — Hyperbola
x²/16 – y²/9 = 1.
Answer
a²=16, b²=9, c²=25, c=5. Foci: (±5,0). Vertices: (±4,0). e=5/4. LR=2(9)/4=9/2.
y²/9 – x²/27 = 1.
Answer
a²=9 (y-axis), b²=27, c²=36, c=6. Foci: (0,±6). Vertices: (0,±3). e=6/3=2. LR=2(27)/3=18.
9y² – 4x² = 36 → y²/4 – x²/9 = 1.
Answer
a²=4, b²=9, c²=13, c=√13. Foci: (0,±√13). Vertices: (0,±2). e=√13/2. LR=2(9)/2=9.
16x² – 9y² = 576 → x²/36 – y²/64 = 1.
Answer
a²=36, b²=64, c²=100, c=10. Foci: (±10,0). Vertices: (±6,0). e=10/6=5/3. LR=2(64)/6=64/3.
5y² – 9x² = 36 → y²/(36/5) – x²/4 = 1.
Answer
a²=36/5, b²=4, c²=36/5+4=56/5. Foci: (0,±√(56/5)). LR=2(4)/√(36/5)=8√5/6=4√5/3.
49y² – 16x² = 784 → y²/16 – x²/49 = 1.
Answer
a²=16, b²=49, c²=65. Foci: (0,±√65). Vertices: (0,±4). e=√65/4. LR=2(49)/4=49/2.
Vertices (±2,0), foci (±3,0).
Answer
a=2, c=3, b²=c²–a²=5. x²/4 – y²/5 = 1
Vertices (0,±5), foci (0,±8).
Answer
a=5, c=8, b²=39. y²/25 – x²/39 = 1
Vertices (0,±3), foci (0,±5).
Answer
a=3, c=5, b²=16. y²/9 – x²/16 = 1
Foci (±5,0), transverse axis length 8.
Answer
2a=8 → a=4, c=5, b²=9. x²/16 – y²/9 = 1
Foci (0,±13), conjugate axis length 24.
Answer
2b=24
⇒ b=12, c=13, a²=169–144=25. y²/25 – x²/144 = 1
Foci (±3√5,0), latus rectum = 8.
Answer
c=3√5, LR=2b²/a=8
⇒ b²=4a. c²=a²+b²: 45=a²+4a
⇒ a²+4a–45=0
⇒ (a+9)(a–5)=0
⇒ a=5.
b²=20. x²/25 – y²/20 = 1
Foci (±4,0), latus rectum = 12.
Answer
c=4, LR=2b²/a=12
⇒ b²=6a. c²=a²+b²: 16=a²+6a
⇒ a²+6a–16=0
⇒ (a+8)(a–2)=0
⇒ a=2.
b²=12. x²/4 – y²/12 = 1
Vertices (±7,0), e=4/3.
Answer
a=7, e=c/a=4/3
⇒ c=28/3. b²=c²–a²=784/9–49=784/9–441/9=343/9.
x²/49 – 9y²/343 = 1
Foci (0,±√10), passing through (2,3).
Answer
c=√10. y²/a² – x²/b² = 1 with b²=c²–a²=10–a².
9/a² – 4/(10–a²) = 1. 9(10–a²)–4a² = a²(10–a²)
90–13a² = 10a²–a⁴
⇒ a⁴–23a²+90=0
⇒ (a²–18)(a²–5)=0
⇒ a²=5 (since b²=10–a²>0)
b²=5. y²/5 – x²/5 = 1 → y²–x²=5
Miscellaneous Exercise
Parabolic reflector 20 cm diameter, 5 cm deep. Find the focus.
Answer
Place vertex at origin with axis along x. The parabola x² = 4ay passes through (10,5):
Wait — diameter is 20cm (radius 10), depth 5cm. Actually let's use y²=4ax form: passes through (5,10).
100 = 4a(5)
⇒ a=5. Focus is at (5,0) i.e., 5 cm from vertex.
Arch: parabola with vertical axis, 10m high, 5m wide at base. Width 2m from vertex.
Answer
Vertex at top. x²=–4ay (opens down). At base: (2.5,–10). 6.25=40a
⇒ a=6.25/40=5/32.
At y=–2: x²=4(5/32)(2)=10/8=5/4
⇒ x=√5/2. Width=2x=√5 m ≈ 2.23 m.
Suspension bridge cable (parabola): 100m span, longest wire 30m, shortest 6m. Find wire at 18m from middle.
Answer
Place vertex at bottom of shortest wire. x²=4ay. At edge (50m from middle, height = 30–6=24m): 2500=4a(24)
⇒ a=2500/96=625/24.
At x=18: y=18²×24/(4×625)=7776/2500=3.1104m. Wire length=6+3.11=9.11m ≈ 9.11m
Semi-ellipse arch: 8m wide, 2m high. Height at 1.5m from one end.
Answer
Ellipse with a=4 (semi-width), b=2 (height). x²/16+y²/4=1.
1.5m from end = 4–1.5=2.5m from centre. At x=2.5: y²=4(1–6.25/16)=4(9.75/16)=2.4375
y=√2.4375 ≈ 1.56m
Rod of length 12cm with ends on axes. Point P is 3cm from end on x-axis. Find locus.
Answer
Let A on x-axis, B on y-axis. AP=3, PB=9. Let angle with x-axis = θ.
x=AP cosθ=3cosθ
⇒ cosθ=x/3; y=PB sinθ=9sinθ
⇒ sinθ=y/9
cos²θ+sin²θ=1
⇒ x²/9 + y²/81 = 1. This is an ellipse.
Area of triangle formed by vertex of x²=12y and ends of its latus rectum.
Answer
x²=12y
⇒ 4a=12
⇒ a=3. Focus: (0,3). LR endpoints: x²=12(3)=36
⇒ x=±6. Points: (6,3) and (–6,3).
Vertex: (0,0). Base=12, height=3. Area = (1/2)(12)(3) = 18 sq units.
Man runs on elliptical track; sum of distances from two flag posts = 10m, distance between posts = 8m. Find equation.
Answer
2a=10 → a=5; 2c=8 → c=4; b²=25–16=9.
Equation: x²/25 + y²/9 = 1
Equilateral triangle inscribed in y²=4ax with one vertex at origin. Find side length.
Answer
Let vertex at origin (0,0). Other two vertices symmetric about x-axis at (at², 2at).
By equilateral triangle condition, all sides equal. Side from origin to point on parabola = √(a²t⁴+4a²t²) = at√(t²+4).
The two symmetric points are (at², 2at) and (at², –2at). Side between them = 4at.
For equilateral: at√(t²+4) = 4at
⇒ √(t²+4)=4
⇒ t²=12
⇒ t=2√3.
Side length = 4at = 4a·2√3 = 8a√3
Important Extra Questions
Find equation of circle passing through (1,2), (3,4) and (5,–6).
Answer
x²+y²+Dx+Ey+F=0. Substituting three points:
1+4+D+2E+F=0 → D+2E+F=–5 ...(1)
9+16+3D+4E+F=0 → 3D+4E+F=–25 ...(2)
25+36+5D–6E+F=0 → 5D–6E+F=–61 ...(3)
(2)–(1): 2D+2E=–20 → D+E=–10 ...(4)
(3)–(2): 2D–10E=–36 → D–5E=–18 ...(5)
(4)–(5): 6E=8
⇒ E=4/3; D=–10–4/3=–34/3; F=–5+34/3–8/3=–5+26/3=11/3
x²+y²–(34/3)x+(4/3)y+11/3=0; multiply by 3: 3x²+3y²–34x+4y+11=0
Find the length of latus rectum of ellipse 3x²+y²=12.
Answer
x²/4+y²/12=1. a²=12, b²=4. LR=2b²/a=2(4)/√12=8/(2√3)=4/√3=4√3/3.
For parabola y²=8x, find point on it such that distance from focus is 6.
Answer
a=2. Distance from focus = x+a = x+2 = 6
⇒ x=4. y²=32
⇒ y=±4√2.
Points: (4, 4√2) and (4, –4√2).
Find eccentricity of ellipse if latus rectum = half of major axis.
Answer
LR = 2b²/a = a
⇒ b²=a²/2. c²=a²–b²=a²/2. e=c/a=1/√2=√2/2.
Find equation of hyperbola with vertices (±3,0) and eccentricity 2.
Answer
a=3, e=c/a=2
⇒ c=6, b²=c²–a²=27. x²/9 – y²/27 = 1