Question 1
Find the circle with centre (0,2) and radius 2.
Steps only
Use (x−h)²+(y−k)²=r².
h=0,k=2,r=2.
Answer
x²+(y−2)²=4.
CLASS 11 · MATHEMATICS · CHAPTER 10
Conic Sections
Work with circles, parabolas, ellipses and hyperbolas using their standard equations. Each answer starts with a compact calculation and opens to a plain-language explanation.
70 worked answersAll 4 exercise setsAll 8 end-of-chapter questions
What you will learn
- Recognise circle equations and find their centres and radii.
- Use focus, vertex and directrix to describe parabolas.
- Find ellipse and hyperbola features from standard forms.
- Model real arches, cables and reflectors with conic equations.
70 worked answers · short steps first, explanations when you need them
Exercise 10.1
Textbook pages 181–182
Question 2
Find the circle with centre (−2,3) and radius 4.
Steps only
h=−2,k=3,r=4.
Substitute into the standard form.
Answer
(x+2)²+(y−3)²=16.
Question 3
Find the circle with centre (1/2,1/4) and radius 1/12.
Steps only
Square the radius: (1/12)²=1/144.
Insert h=1/2,k=1/4.
Answer
(x−1/2)²+(y−1/4)²=1/144.
Question 4
Find the circle with centre (1,1) and radius 2.
Steps only
Use h=1,k=1,r=2.
Answer
(x−1)²+(y−1)²=4.
Question 5
Find the circle with centre (−a,−b) and radius √(a²−b²).
Steps only
Centre coordinates are h=−a,k=−b.
The radius squared is a²−b².
Answer
(x+a)²+(y+b)²=a²−b².
Question 6
Find the centre and radius of (x+5)²+(y−3)²=36.
Steps only
Compare with (x−h)²+(y−k)²=r².
r=√36=6.
Answer
Centre (−5,3); radius 6.
Question 7
Find centre and radius of x²+y²−4x−8y−45=0.
Steps only
Complete squares: (x−2)²+(y−4)²=65.
r=√65.
Answer
Centre (2,4); radius √65.
Question 8
Find centre and radius of x²+y²−8x+10y−12=0.
Steps only
(x−4)²−16+(y+5)²−25−12=0.
(x−4)²+(y+5)²=53.
Answer
Centre (4,−5); radius √53.
Question 9
Find centre and radius of 2x²+2y²−x=0.
Steps only
Divide by 2: x²+y²−x/2=0.
(x−1/4)²+y²=1/16.
Answer
Centre (1/4,0); radius 1/4.
Question 10
Find the circle through (4,1),(6,5) whose centre lies on 4x+y=16.
Steps only
The centre is equidistant from the two points and satisfies 4h+k=16.
Solving gives (h,k)=(3,4).
Radius squared=(4−3)²+(1−4)²=10.
Answer
(x−3)²+(y−4)²=10.
Question 11
Find the circle through (2,3),(−1,1) with centre on x−3y−11=0.
Steps only
Equidistance gives 6h+4k=11; centre condition h−3k=11.
Solving gives h=7/2,k=−5/2.
r²=(7/2−2)²+(−5/2−3)²=65/2.
Answer
(x−7/2)²+(y+5/2)²=65/2.
Question 12
Find circles of radius 5 with centre on the x-axis passing through (2,3).
Steps only
Let centre be (h,0).
(2−h)²+3²=25 gives h=6 or h=−2.
Answer
(x−6)²+y²=25 or (x+2)²+y²=25.
Question 13
Find the circle through (0,0) with non-zero axis intercepts a and b.
Steps only
Use x²+y²+Dx+Ey+F=0 and the origin gives F=0.
At (a,0): D=−a; at (0,b): E=−b.
Answer
x²+y²−ax−by=0.
Question 14
Find the circle with centre (2,2) passing through (4,5).
Steps only
r²=(4−2)²+(5−2)²=4+9=13.
Answer
(x−2)²+(y−2)²=13.
Question 15
Does (−2.5,3.5) lie inside, outside or on x²+y²=25?
Steps only
(−2.5)²+(3.5)²=6.25+12.25=18.5.
18.5<25.
Answer
Inside.
Exercise 10.2
Textbook pages 186–187
Question 1
For y²=12x, find focus, axis, directrix and latus rectum length.
Steps only
y²=4ax with 4a=12, so a=3.
For a standard parabola, the latus rectum length is 4a.
Answer
Focus (3,0); axis x-axis; directrix x=−3; latus rectum 12.
Question 2
For x²=6y, find focus, axis, directrix and latus rectum length.
Steps only
x²=4ay with 4a=6, so a=3/2.
For a standard parabola, the latus rectum length is 4a.
Answer
Focus (0,3/2); axis y-axis; directrix y=−3/2; latus rectum 6.
Question 3
For y²=−8x, find focus, axis, directrix and latus rectum length.
Steps only
Compare with y²=−4ax: a=2.
For a standard parabola, the latus rectum length is 4a.
Answer
Focus (−2,0); axis x-axis; directrix x=2; latus rectum 8.
Question 4
For x²=−16y, find focus, axis, directrix and latus rectum length.
Steps only
Compare with x²=−4ay: a=4.
For a standard parabola, the latus rectum length is 4a.
Answer
Focus (0,−4); axis y-axis; directrix y=4; latus rectum 16.
Question 5
For y²=10x, find focus, axis, directrix and latus rectum length.
Steps only
Here 4a=10, so a=5/2.
For a standard parabola, the latus rectum length is 4a.
Answer
Focus (5/2,0); axis x-axis; directrix x=−5/2; latus rectum 10.
Question 6
For x²=−9y, find focus, axis, directrix and latus rectum length.
Steps only
Compare with x²=−4ay, giving a=9/4.
For a standard parabola, the latus rectum length is 4a.
Answer
Focus (0,−9/4); axis y-axis; directrix y=9/4; latus rectum 9.
Question 7
Find the parabola with focus (6,0) and directrix x=−6.
Steps only
The vertex is midway between focus and directrix, at the origin.
a=6; y²=4ax.
Answer
y²=24x.
Question 8
Find the parabola with focus (0,−3) and directrix y=3.
Steps only
Vertex is (0,0), a=3 and it opens down.
x²=−4ay.
Answer
x²=−12y.
Question 9
Find the parabola with vertex (0,0) and focus (3,0).
Steps only
a=3 and the axis is the x-axis.
Use y²=4ax.
Answer
y²=12x.
Question 10
Find the parabola with vertex (0,0) and focus (−2,0).
Steps only
a=2 and it opens left.
Use y²=−4ax.
Answer
y²=−8x.
Question 11
A parabola has vertex (0,0), axis along x-axis, and passes through (2,3). Find its equation.
Steps only
Use y²=4ax.
9=4a(2), so a=9/8.
Answer
y²=(9/2)x.
Question 12
A parabola has vertex (0,0), is symmetric about the y-axis, and passes through (5,2). Find its equation.
Steps only
Use x²=4ay.
25=4a(2), so a=25/8.
Answer
x²=(25/2)y.
Exercise 10.3
Textbook pages 195–196
Question 1
For ellipse x²/36+y²/16=1, find foci, vertices, axis lengths, eccentricity and latus rectum.
Steps only
a=6,b=4,c=√(a²−b²)=2√5.
Use c²=a²−b², e=c/a and latus rectum 2b²/a.
Answer
Foci (±2√5,0); vertices (±6,0); minor endpoints (0,±4); axes 12,8; e=√5/3; latus rectum=16/3.
Question 2
For ellipse x²/4+y²/25=1, find foci, vertices, axis lengths, eccentricity and latus rectum.
Steps only
Major axis is y; a=5,b=2,c=√21.
Use c²=a²−b², e=c/a and latus rectum 2b²/a.
Answer
Foci (0,±√21); vertices (0,±5); minor endpoints (±2,0); axes 10,4; e=√21/5; latus rectum=8/5.
Question 3
For ellipse x²/16+y²/9=1, find foci, vertices, axis lengths, eccentricity and latus rectum.
Steps only
a=4,b=3,c=√7.
Use c²=a²−b², e=c/a and latus rectum 2b²/a.
Answer
Foci (±√7,0); vertices (±4,0); minor endpoints (0,±3); axes 8,6; e=√7/4; latus rectum=9/2.
Question 4
For ellipse x²/25+y²/100=1, find foci, vertices, axis lengths, eccentricity and latus rectum.
Steps only
Major axis is y; a=10,b=5,c=√75=5√3.
Use c²=a²−b², e=c/a and latus rectum 2b²/a.
Answer
Foci (0,±5√3); vertices (0,±10); minor endpoints (±5,0); axes 20,10; e=√3/2; latus rectum=5.
Question 5
For ellipse x²/49+y²/36=1, find foci, vertices, axis lengths, eccentricity and latus rectum.
Steps only
a=7,b=6,c=√13.
Use c²=a²−b², e=c/a and latus rectum 2b²/a.
Answer
Foci (±√13,0); vertices (±7,0); minor endpoints (0,±6); axes 14,12; e=√13/7; latus rectum=72/7.
Question 6
For ellipse x²/100+y²/400=1, find foci, vertices, axis lengths, eccentricity and latus rectum.
Steps only
Major axis is y; a=20,b=10,c=√300=10√3.
Use c²=a²−b², e=c/a and latus rectum 2b²/a.
Answer
Foci (0,±10√3); vertices (0,±20); minor endpoints (±10,0); axes 40,20; e=√3/2; latus rectum=10.
Question 7
For ellipse 36x²+4y²=144, find foci, vertices, axis lengths, eccentricity and latus rectum.
Steps only
Divide by 144: x²/4+y²/36=1, so a=6,b=2,c=4√2.
Use c²=a²−b², e=c/a and latus rectum 2b²/a.
Answer
Foci (0,±4√2); vertices (0,±6); minor endpoints (±2,0); axes 12,4; e=2√2/3; latus rectum=4/3.
Question 8
For ellipse 16x²+y²=16, find foci, vertices, axis lengths, eccentricity and latus rectum.
Steps only
Divide by 16: x²/1+y²/16=1; a=4,b=1,c=√15.
Use c²=a²−b², e=c/a and latus rectum 2b²/a.
Answer
Foci (0,±√15); vertices (0,±4); minor endpoints (±1,0); axes 8,2; e=√15/4; latus rectum=1/2.
Question 9
For ellipse 4x²+9y²=36, find foci, vertices, axis lengths, eccentricity and latus rectum.
Steps only
Divide by 36 to get x²/9+y²/4=1.
Use c²=a²−b², e=c/a and latus rectum 2b²/a.
Answer
Foci (±√5,0); vertices (±3,0); minor endpoints (0,±2); axes 6,4; e=√5/3; latus rectum=8/3.
Question 10
Find the ellipse: Vertices (±5,0), foci (±4,0).
Steps only
a=5,c=4,b²=25−16=9.
Place the larger semi-axis along the stated major-axis direction.
Answer
x²/25+y²/9=1.
Question 11
Find the ellipse: Vertices (0,±13), foci (0,±5).
Steps only
a=13,c=5,b²=169−25=144.
Place the larger semi-axis along the stated major-axis direction.
Answer
x²/144+y²/169=1.
Question 12
Find the ellipse: Vertices (±6,0), foci (±4,0).
Steps only
a=6,c=4,b²=36−16=20.
Place the larger semi-axis along the stated major-axis direction.
Answer
x²/36+y²/20=1.
Question 13
Find the ellipse: Major-axis endpoints (±3,0), minor-axis endpoints (0,±2).
Steps only
The semi-axes are a=3 and b=2.
Place the larger semi-axis along the stated major-axis direction.
Answer
x²/9+y²/4=1.
Question 14
Find the ellipse: Major-axis endpoints (0,±√5), minor-axis endpoints (±1,0).
Steps only
The semi-major axis is √5 along y; the semi-minor axis is 1 along x.
Place the larger semi-axis along the stated major-axis direction.
Answer
x²+y²/5=1.
Question 15
Find the ellipse: Major-axis length 26; foci (±5,0).
Steps only
a=13,c=5,b²=169−25=144.
Place the larger semi-axis along the stated major-axis direction.
Answer
x²/169+y²/144=1.
Question 16
Find the ellipse: Minor-axis length 16; foci (0,±6).
Steps only
b=8,c=6,a²=64+36=100; major axis is y.
Place the larger semi-axis along the stated major-axis direction.
Answer
x²/64+y²/100=1.
Question 17
Find the ellipse: Foci (±3,0) and a=4.
Steps only
c=3; b²=a²−c²=16−9=7.
Place the larger semi-axis along the stated major-axis direction.
Answer
x²/16+y²/7=1.
Question 18
Find the ellipse: b=3,c=4, centre at origin and foci on x-axis.
Steps only
a²=b²+c²=9+16=25.
Place the larger semi-axis along the stated major-axis direction.
Answer
x²/25+y²/9=1.
Question 19
Find the ellipse: Centre at origin, major axis on y-axis, passing through (3,2) and (1,6).
Steps only
Let u=1/b²,v=1/a². The points give 9u+4v=1 and u+36v=1, so u=1/10,v=1/40.
Place the larger semi-axis along the stated major-axis direction.
Answer
x²/10+y²/40=1.
Question 20
Find the ellipse: Ellipse with major axis on x-axis passing through (4,3) and (6,2).
Steps only
Let u=1/a²,v=1/b². Then 16u+9v=1 and 36u+4v=1, giving u=1/52,v=1/13.
Place the larger semi-axis along the stated major-axis direction.
Answer
x²/52+y²/13=1.
Exercise 10.4
Textbook pages 202
Question 1
For hyperbola x²/16−y²/9=1, find foci, vertices, eccentricity and latus rectum.
Steps only
a=4,b=3,c=√(a²+b²)=5.
Use c²=a²+b², e=c/a and latus rectum 2b²/a.
Answer
Foci (±5,0); vertices (±4,0); e=5/4; latus rectum=9/2.
Question 2
For hyperbola y²/9−x²/27=1, find foci, vertices, eccentricity and latus rectum.
Steps only
a=3,b=3√3,c=√(a²+b²)=6.
Use c²=a²+b², e=c/a and latus rectum 2b²/a.
Answer
Foci (0,±6); vertices (0,±3); e=2; latus rectum=18.
Question 3
For hyperbola 9y²−4x²=36, find foci, vertices, eccentricity and latus rectum.
Steps only
Divide by 36: y²/4−x²/9=1; a=2,b=3,c=√13.
Use c²=a²+b², e=c/a and latus rectum 2b²/a.
Answer
Foci (0,±√13); vertices (0,±2); e=√13/2; latus rectum=9.
Question 4
For hyperbola 16x²−9y²=576, find foci, vertices, eccentricity and latus rectum.
Steps only
Divide by 576: x²/36−y²/64=1; a=6,b=8,c=10.
Use c²=a²+b², e=c/a and latus rectum 2b²/a.
Answer
Foci (±10,0); vertices (±6,0); e=5/3; latus rectum=64/3.
Question 5
For hyperbola 5y²−9x²=36, find foci, vertices, eccentricity and latus rectum.
Steps only
Divide by 36: y²/(36/5)−x²/4=1; a=6/√5,b=2,c=2√70/5.
Use c²=a²+b², e=c/a and latus rectum 2b²/a.
Answer
Foci (0,±2√70/5); vertices (0,±6/√5); e=√14/3; latus rectum=4√5/3.
Question 6
For hyperbola 49y²−16x²=784, find foci, vertices, eccentricity and latus rectum.
Steps only
Divide by 784: y²/16−x²/49=1; a=4,b=7,c=√65.
Use c²=a²+b², e=c/a and latus rectum 2b²/a.
Answer
Foci (0,±√65); vertices (0,±4); e=√65/4; latus rectum=49/2.
Question 7
Find the hyperbola: Vertices (±2,0), foci (±3,0).
Steps only
a=2,c=3,b²=c²−a²=5.
Use c²=a²+b² and the specified axis orientation.
Answer
x²/4−y²/5=1.
Question 8
Find the hyperbola: Vertices (0,±5), foci (0,±8).
Steps only
a=5,c=8,b²=64−25=39.
Use c²=a²+b² and the specified axis orientation.
Answer
y²/25−x²/39=1.
Question 9
Find the hyperbola: Vertices (0,±3), foci (0,±5).
Steps only
a=3,c=5,b²=25−9=16.
Use c²=a²+b² and the specified axis orientation.
Answer
y²/9−x²/16=1.
Question 10
Find the hyperbola: Foci (±5,0), transverse-axis length 8.
Steps only
c=5,a=4,b²=c²−a²=9.
Use c²=a²+b² and the specified axis orientation.
Answer
x²/16−y²/9=1.
Question 11
Find the hyperbola: Foci (0,±13), conjugate-axis length 24.
Steps only
c=13,b=12,a²=c²−b²=25.
Use c²=a²+b² and the specified axis orientation.
Answer
y²/25−x²/144=1.
Question 12
Find the hyperbola: Foci (±3√5,0), latus rectum length 8.
Steps only
c²=45 and 2b²/a=8, so b²=4a. Solve a²+4a=45: a=5,b²=20.
Use c²=a²+b² and the specified axis orientation.
Answer
x²/25−y²/20=1.
Question 13
Find the hyperbola: Foci (±4,0), latus rectum length 12.
Steps only
c=4 and b²=6a. From a²+b²=16, a²+6a−16=0 gives a=2,b²=12.
Use c²=a²+b² and the specified axis orientation.
Answer
x²/4−y²/12=1.
Question 14
Find the hyperbola: Vertices (±7,0), eccentricity 4/3.
Steps only
a=7,c=ae=28/3; b²=c²−a²=343/9.
Use c²=a²+b² and the specified axis orientation.
Answer
x²/49−y²/(343/9)=1.
Question 15
Find the hyperbola: Foci (0,±√10), passing through (2,3).
Steps only
Let the equation be y²/A−x²/B=1. Since A+B=10 and (2,3) lies on it, 9/A−4/B=1; solving gives A=B=5.
Use c²=a²+b² and the specified axis orientation.
Answer
y²/5−x²/5=1.
Miscellaneous Exercise on Chapter 10
Textbook pages 202–204
Question 1
A parabolic reflector is 20 cm across and 5 cm deep. Find its focus.
Steps only
Take the vertex as origin and the axis along x.
At the rim, x=5 and y=±10; y²=4ax gives 100=20a.
a=5, so focus is (5,0).
Answer
5 cm from the vertex.
Question 2
A vertical-axis parabolic arch is 10 m high and 5 m wide at its base. Find its width 2 m below the vertex.
Steps only
Set vertex at (0,10); model x²=4a(y−10).
At the base, x=5/2,y=0 gives a=−5/32.
At y=8, x²=5/4, so full width=2|x|=√5.
Answer
√5 m.
Question 3
A 100 m suspension-bridge roadway has supporting wires from a parabolic cable. The centre wire is 6 m and end wires 30 m. Find the wire 18 m from the middle.
Steps only
Let cable height be y=6+ax².
At x=50, 30=6+2500a, so a=0.0096.
At x=18, y=6+0.0096(324)=9.1104.
Answer
9.1104 m.
Question 4
A semi-elliptical arch is 8 m wide and 2 m high at its centre. Find its height 1.5 m from an end.
Steps only
Semi-axes are a=4,b=2: x²/16+y²/4=1.
The point is x=4−1.5=2.5 from the centre.
y²=39/16, so y=√39/4.
Answer
√39/4 m (about 1.56 m).
Question 5
A 12 cm rod has ends on the coordinate axes. Find the locus of a point 3 cm from the end on the x-axis.
Steps only
Let rod endpoints be (a,0),(0,b), with a²+b²=144.
The point 3/12 of the way from (a,0) is (3a/4,b/4).
Thus a=4x/3,b=4y and x²/81+y²/9=1.
Answer
x²/81+y²/9=1.
Question 6
Find the area of the triangle from the vertex of x²=12y to the ends of its latus rectum.
Steps only
Here 4a=12, so a=3 and the vertex is (0,0).
Latus-rectum endpoints are (−6,3),(6,3).
Base=12, height=3; area=18.
Answer
18 square units.
Question 7
A runner stays so the sum of distances from two flag posts is 10 m; the posts are 8 m apart. Find the runner’s locus.
Steps only
For an ellipse, 2a=10 so a=5; focal separation 2c=8 so c=4.
b²=a²−c²=25−16=9.
Use x²/a²+y²/b²=1.
Answer
x²/25+y²/9=1, with the posts at (±4,0).
Question 8
An equilateral triangle is inscribed in y²=4ax with one vertex at the parabola’s vertex. Find its side length.
Steps only
By symmetry let the other vertices be (x,y),(x,−y).
Equal side lengths give x²+y²=(2y)², hence x²=3y².
Using y²=4ax gives y=4√3a; side=2y=8√3a.
Answer
8√3a.