SUMMARY: The chapter on Conic Sections in Class 11 Mathematics explores the definitions, properties, and equations of various conic sections such as circles, ellipses, parabolas, and hyperbolas. KEY TOPICS: conic sections, circle, ellipse, parabola, hyperbola, standard equations, eccentricity, directrix, focus, latus rectum
Find the equation of the circle passing through (0, 0), (4, 0) and (0, 4).
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General circle: x² + y² + Dx + Ey + F = 0. Through (0, 0): F = 0. Through (4, 0): 16 + 4D = 0 ⇒ D = −4. Through (0, 4): 16 + 4E = 0 ⇒ E = −4. Equation: x² + y² − 4x − 4y = 0, i.e. (x − 2)² + (y − 2)² = 8 with centre (2, 2) and radius 2√2.
Q126 Marks
Find the equation of the parabola with vertex at origin, axis along x-axis and passing through the point (3, 6).
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Parabola y² = 4ax. Through (3, 6): 36 = 4a · 3 ⇒ a = 3. Equation y² = 12x.
Q136 Marks
For the ellipse x²/25 + y²/16 = 1 find the lengths of major and minor axes the foci and the eccentricity.
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a² = 25 ⇒ a = 5; b² = 16 ⇒ b = 4. Major axis = 2a = 10; minor axis = 2b = 8. c² = a² − b² = 9 ⇒ c = 3. Foci: (±3, 0). Eccentricity e = c/a = 3/5.
Q146 Marks
Find the equation of the hyperbola with foci (±5, 0) and length of conjugate axis 8.
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Foci (±c, 0) so c = 5. Conjugate axis 2b = 8 so b = 4. Then a² = c² − b² = 25 − 16 = 9 ⇒ a = 3. Equation: x²/9 − y²/16 = 1.
Q156 Marks
For the parabola y² = 8x, find the length of latus rectum and the focal distance from the point (2, 4) to the focus.
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y² = 4ax with 4a = 8 so a = 2. Latus rectum = 4a = 8. Focus is (2, 0). Distance from (2, 4) to focus (2, 0) = √(0² + 4²) = 4 units.
Q166 Marks
Compare ellipse hyperbola and parabola with the help of a table.
Assertion–Reason Questions5 questions
Q171 Mark
Assertion (A): A circle is a special case of an ellipse.
Reason (R): A circle is an ellipse with equal semi-axes (eccentricity 0).
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Correct answer: Option 1 —
Both A and R are true, and R is the correct explanation of A.
Q181 Mark
Assertion (A): The axis of the parabola y² = 4ax is the x-axis.
Reason (R): The parabola is symmetric about the x-axis.
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Correct answer: Option 1 —
Both A and R are true, and R is the correct explanation of A.
Q191 Mark
Assertion (A): For an ellipse a² > b² and e < 1.
Reason (R): The ellipse equation is x²/a² + y²/b² = 1 with semi-major axis a and semi-minor axis b.
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Correct answer: Option 1 —
Both A and R are true, and R is the correct explanation of A.
Q201 Mark
Assertion (A): The eccentricity of a hyperbola is greater than 1.
Reason (R): For a hyperbola c > a so e = c/a > 1.
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Correct answer: Option 1 —
Both A and R are true, and R is the correct explanation of A.
Q211 Mark
Assertion (A): The equation x² + y² = 4 represents a circle of radius 2.
Reason (R): Comparing with x² + y² = r² gives r² = 4 ⇒ r = 2.
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Correct answer: Option 1 —
Both A and R are true, and R is the correct explanation of A.
Statement-Based Questions5 questions
Q221 Mark
Statement 1: A circle is a conic section with eccentricity 0.
Statement 2: A parabola has eccentricity 1.
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Correct answer: Option 1 —
Both statements are true.
Q231 Mark
Statement 1: An ellipse has two foci.
Statement 2: The sum of distances from any point on the ellipse to the foci is constant.
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Correct answer: Option 1 —
Both statements are true.
Q241 Mark
Statement 1: A hyperbola has two branches.
Statement 2: The difference of distances from any point on the hyperbola to the two foci is constant.
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Correct answer: Option 1 —
Both statements are true.
Q251 Mark
Statement 1: The parabola y² = 4ax opens to the right when a > 0.
Statement 2: The parabola y² = 4ax opens to the left when a < 0.
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Correct answer: Option 1 —
Both statements are true.
Q261 Mark
Statement 1: A parabola is the locus of points equidistant from a focus and a directrix.
Statement 2: The eccentricity of a parabola is 1.
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Correct answer: Option 1 —
Both statements are true.
Case Study / Passage Questions3 questions
Q273 Marks
A satellite dish has a parabolic cross-section with the vertex at the origin opening to the right described by y² = 4ax where a is the focal distance. The receiver is placed at the focus 30 cm from the vertex.
If a = 30 cm the equation of the parabola is:
Ay² = 30x
By² = 60x
Cy² = 90x
Dy² = 120x
The length of the latus rectum equals:
A30 cm
B60 cm
C120 cm
D150 cm
Find a point on the parabola at horizontal distance 50 cm from the vertex.
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1. Option 4 — y² = 120x
2. Option 3 — 120 cm
3. With a = 30: equation y² = 4ax = 120x. Latus rectum = 4a = 120 cm. Focus is at (30, 0) and the directrix is x = −30. The vertex (0, 0) is on the parabola.
Q283 Marks
An elliptical garden has the equation x²/100 + y²/64 = 1 (in metres). The gardener wants to know the major axis minor axis foci and eccentricity.
The semi-major axis equals:
A8 m
B10 m
C16 m
D20 m
The eccentricity equals:
A3/5
B4/5
C5/3
D5/4
Find the foci of the ellipse.
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1. Option 2 — 10 m
2. Option 1 — 3/5
3. a² = 100 ⇒ a = 10 (semi-major). b² = 64 ⇒ b = 8 (semi-minor). c = √(a² − b²) = √36 = 6. Eccentricity = c/a = 6/10 = 3/5. Foci at (±6, 0). Major axis = 2a = 20 m; minor axis = 2b = 16 m.
Q293 Marks
A hyperbolic mirror has the equation x²/9 − y²/16 = 1. The optician needs the foci eccentricity asymptotes and the directrix lines.
The semi-transverse axis 'a' equals:
A3
B4
C5
D7
The eccentricity equals:
A3/5
B4/3
C5/3
D5/4
Write the equations of the asymptotes.
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1. Option 1 — 3
2. Option 3 — 5/3
3. a² = 9 ⇒ a = 3; b² = 16 ⇒ b = 4. c² = a² + b² = 25 ⇒ c = 5. e = c/a = 5/3 > 1 ✓ (hyperbola). Foci at (±5, 0). Asymptotes y = ±(b/a) x = ±(4/3) x.
Table-Based Questions3 questions
Q303 Marks
Study the standard equations of conics:
Conic
Standard equation
Eccentricity
Circle
x² + y² = r²
0
Parabola
y² = 4ax
1
Ellipse
x²/a² + y²/b² = 1
< 1
Hyperbola
x²/a² − y²/b² = 1
> 1
The eccentricity of a circle is:
A0
B1
C>1
D<1
The conic with eccentricity equal to 1 is the:
ACircle
BParabola
CEllipse
DHyperbola
Write the standard equation and eccentricity range for each conic.
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1. Option 1 — 0
2. Option 2 — Parabola
3. Eccentricity classifies conics: e = 0 → circle; e = 1 → parabola; 0 < e < 1 → ellipse; e > 1 → hyperbola. As e increases the conic 'opens up' more.
Q313 Marks
Study the parabola y² = 12x and its parameters:
Parameter
Value
4a
12
a
3
Vertex
(0, 0)
Focus
(3, 0)
Directrix
x = −3
Latus rectum
12
The value of a (focal distance) is:
A1
B3
C6
D12
The length of the latus rectum equals:
A3
B6
C12
D24
Find the equation of a chord of length 8 perpendicular to the axis at x = 3.
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1. Option 2 — 3
2. Option 3 — 12
3. For y² = 4ax: 4a = 12 ⇒ a = 3. Vertex (0, 0). Focus (a, 0) = (3, 0). Directrix x = −a = −3. Latus rectum = 4a = 12. The parabola opens to the right since a > 0.
Q326 Marks
For the ellipse x²/25 + y²/9 = 1, find the lengths of major and minor axes, the foci, and the eccentricity.
Parameter
Computation
a²
25
b²
9
c² = a² − b²
?
Eccentricity e = c/a
?
Picture-Based Questions1 question
Q333 Marks
Study the four conic sections plotted together and answer:
The conic with eccentricity e = 0 is the:
ACircle
BParabola
CEllipse
DHyperbola
The eccentricity of a hyperbola is:
A0
B< 1
C1
D> 1
Classify the four conics by their eccentricity.
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1. Option 1 — Circle
2. Option 4 — > 1
3. Eccentricity e classifies conic sections: e = 0 → circle (special ellipse); 0 < e < 1 → ellipse; e = 1 → parabola; e > 1 → hyperbola. As e increases, the curve becomes more 'open'. All four conics arise from cutting a double cone with a plane at different angles.
