Exercise 10.3 answers: Conic Sections

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1. Compare with the standard form x²/a² + y²/b² = 1: here a² = 36 and b² = 16.
2. Since 36 > 16, the major axis lies along the x-axis, with a = 6 and b = 4.
3. c² = a² − b² = 36 − 16 = 20, so c = 2√5.
4. Foci are at (±c, 0) = (±2√5, 0); vertices are at (±a, 0) = (±6, 0).
5. Length of major axis = 2a = 12; length of minor axis = 2b = 8.
6. Eccentricity e = c/a = 2√5/6 = √5/3.
7. Length of latus rectum = 2b²/a = 2(16)/6 = 16/3.

AnswerFoci (±2√5, 0); vertices (±6, 0); major axis 12; minor axis 8; eccentricity √5/3; latus rectum 16/3.

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1. Compare with the standard form: here 25 sits under y² and 4 under x².
2. Since 25 > 4, the major axis lies along the y-axis, with a = 5 and b = 2.
3. c² = a² − b² = 25 − 4 = 21, so c = √21.
4. Foci are at (0, ±c) = (0, ±√21); vertices are at (0, ±a) = (0, ±5).
5. Length of major axis = 2a = 10; length of minor axis = 2b = 4.
6. Eccentricity e = c/a = √21/5.
7. Length of latus rectum = 2b²/a = 2(4)/5 = 8/5.

AnswerFoci (0, ±√21); vertices (0, ±5); major axis 10; minor axis 4; eccentricity √21/5; latus rectum 8/5.

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Open NCERT p. 195Matches NCERT’s answer

1. Compare with the standard form: here a² = 16 and b² = 9.
2. Since 16 > 9, the major axis lies along the x-axis, with a = 4 and b = 3.
3. c² = a² − b² = 16 − 9 = 7, so c = √7.
4. Foci are at (±c, 0) = (±√7, 0); vertices are at (±a, 0) = (±4, 0).
5. Length of major axis = 2a = 8; length of minor axis = 2b = 6.
6. Eccentricity e = c/a = √7/4.
7. Length of latus rectum = 2b²/a = 2(9)/4 = 9/2.

AnswerFoci (±√7, 0); vertices (±4, 0); major axis 8; minor axis 6; eccentricity √7/4; latus rectum 9/2.

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Open NCERT p. 195Matches NCERT’s answer

1. Compare with the standard form: here 100 sits under y² and 25 under x².
2. Since 100 > 25, the major axis lies along the y-axis, with a = 10 and b = 5.
3. c² = a² − b² = 100 − 25 = 75, so c = 5√3.
4. Foci are at (0, ±c) = (0, ±5√3); vertices are at (0, ±a) = (0, ±10).
5. Length of major axis = 2a = 20; length of minor axis = 2b = 10.
6. Eccentricity e = c/a = 5√3/10 = √3/2.
7. Length of latus rectum = 2b²/a = 2(25)/10 = 5.

AnswerFoci (0, ±5√3); vertices (0, ±10); major axis 20; minor axis 10; eccentricity √3/2; latus rectum 5.

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Open NCERT p. 195Matches NCERT’s answer

1. Compare with the standard form: here a² = 49 and b² = 36.
2. Since 49 > 36, the major axis lies along the x-axis, with a = 7 and b = 6.
3. c² = a² − b² = 49 − 36 = 13, so c = √13.
4. Foci are at (±c, 0) = (±√13, 0); vertices are at (±a, 0) = (±7, 0).
5. Length of major axis = 2a = 14; length of minor axis = 2b = 12.
6. Eccentricity e = c/a = √13/7.
7. Length of latus rectum = 2b²/a = 2(36)/7 = 72/7.

AnswerFoci (±√13, 0); vertices (±7, 0); major axis 14; minor axis 12; eccentricity √13/7; latus rectum 72/7.

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Open NCERT p. 195Matches NCERT’s answer

1. Compare with the standard form: here 400 sits under y² and 100 under x².
2. Since 400 > 100, the major axis lies along the y-axis, with a = 20 and b = 10.
3. c² = a² − b² = 400 − 100 = 300, so c = 10√3.
4. Foci are at (0, ±c) = (0, ±10√3); vertices are at (0, ±a) = (0, ±20).
5. Length of major axis = 2a = 40; length of minor axis = 2b = 20.
6. Eccentricity e = c/a = 10√3/20 = √3/2.
7. Length of latus rectum = 2b²/a = 2(100)/20 = 10.

AnswerFoci (0, ±10√3); vertices (0, ±20); major axis 40; minor axis 20; eccentricity √3/2; latus rectum 10.

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1. The equation is not yet in standard form, so divide every term by 144: x²/4 + y²/36 = 1.
2. Here a² = 36 (under y²) and b² = 4 (under x²); since 36 > 4, the major axis lies along the y-axis.
3. So a = 6 and b = 2. c² = a² − b² = 36 − 4 = 32, so c = 4√2.
4. Foci are at (0, ±c) = (0, ±4√2); vertices are at (0, ±a) = (0, ±6).
5. Length of major axis = 2a = 12; length of minor axis = 2b = 4.
6. Eccentricity e = c/a = 4√2/6 = 2√2/3.
7. Length of latus rectum = 2b²/a = 2(4)/6 = 4/3.

AnswerFoci (0, ±4√2); vertices (0, ±6); major axis 12; minor axis 4; eccentricity 2√2/3; latus rectum 4/3.

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Open NCERT p. 195Matches NCERT’s answer

1. The equation is not yet in standard form, so divide every term by 16: x²/1 + y²/16 = 1.
2. Here a² = 16 (under y²) and b² = 1 (under x²); since 16 > 1, the major axis lies along the y-axis.
3. So a = 4 and b = 1. c² = a² − b² = 16 − 1 = 15, so c = √15.
4. Foci are at (0, ±c) = (0, ±√15); vertices are at (0, ±a) = (0, ±4).
5. Length of major axis = 2a = 8; length of minor axis = 2b = 2.
6. Eccentricity e = c/a = √15/4.
7. Length of latus rectum = 2b²/a = 2(1)/4 = 1/2.

AnswerFoci (0, ±√15); vertices (0, ±4); major axis 8; minor axis 2; eccentricity √15/4; latus rectum 1/2.

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Open NCERT p. 195Matches NCERT’s answer

1. The equation is not yet in standard form, so divide every term by 36: x²/9 + y²/4 = 1.
2. Here a² = 9 (under x²) and b² = 4 (under y²); since 9 > 4, the major axis lies along the x-axis.
3. So a = 3 and b = 2. c² = a² − b² = 9 − 4 = 5, so c = √5.
4. Foci are at (±c, 0) = (±√5, 0); vertices are at (±a, 0) = (±3, 0).
5. Length of major axis = 2a = 6; length of minor axis = 2b = 4.
6. Eccentricity e = c/a = √5/3.
7. Length of latus rectum = 2b²/a = 2(4)/3 = 8/3.

AnswerFoci (±√5, 0); vertices (±3, 0); major axis 6; minor axis 4; eccentricity √5/3; latus rectum 8/3.

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1. The vertices and the foci are both on the x-axis, so the major axis lies along the x-axis.
2. Vertices (± 5, 0) give a = 5.
3. Foci (± 4, 0) give c = 4.
4. For an ellipse, b² = a² − c² = 25 − 16 = 9.
5. Write the standard equation x²/a² + y²/b² = 1 with a² = 25, b² = 9.

Answerx²/25 + y²/9 = 1

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Open NCERT p. 195Matches NCERT’s answer

1. The vertices and the foci are on the y-axis, so the major axis lies along the y-axis.
2. Vertices (0, ± 13) give a = 13.
3. Foci (0, ± 5) give c = 5.
4. b² = a² − c² = 169 − 25 = 144.
5. Since the major axis is along y, the bigger denominator goes under y²: x²/b² + y²/a² = 1.

Answerx²/144 + y²/169 = 1

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Open NCERT p. 195Matches NCERT’s answer

1. Vertices and foci are on the x-axis, so the major axis lies along the x-axis.
2. Vertices (± 6, 0) give a = 6.
3. Foci (± 4, 0) give c = 4.
4. b² = a² − c² = 36 − 16 = 20.
5. Write x²/36 + y²/20 = 1.

Answerx²/36 + y²/20 = 1

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1. The ends of the major axis are the vertices, so a = 3.
2. The ends of the minor axis give b = 2.
3. The major axis is along the x-axis, so a² sits under x².
4. Write x²/9 + y²/4 = 1.

Answerx²/9 + y²/4 = 1

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Open NCERT p. 195Matches NCERT’s answer

1. The ends of the major axis are on the y-axis, so a = √5 and the major axis is along y.
2. The ends of the minor axis give b = 1.
3. Since the major axis is along y, a² sits under y²: x²/b² + y²/a² = 1.
4. Write x²/1 + y²/5 = 1.

Answerx² + y²/5 = 1

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1. The length of the major axis is 2a, so 2a = 26 and a = 13.
2. The foci (± 5, 0) are on the x-axis, so c = 5 and the major axis is along x.
3. b² = a² − c² = 169 − 25 = 144.
4. Write x²/169 + y²/144 = 1.

Answerx²/169 + y²/144 = 1

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1. The length of the minor axis is 2b, so 2b = 16 and b = 8.
2. The foci (0, ± 6) are on the y-axis, so c = 6 and the major axis is along y.
3. a² = b² + c² = 64 + 36 = 100.
4. Since the major axis is along y, write x²/64 + y²/100 = 1.

Answerx²/64 + y²/100 = 1

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Open NCERT p. 195Matches NCERT’s answer

1. Foci (± 3, 0) are on the x-axis, so c = 3 and the major axis is along x.
2. a is given as 4.
3. b² = a² − c² = 16 − 9 = 7.
4. Write x²/16 + y²/7 = 1.

Answerx²/16 + y²/7 = 1

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Open NCERT p. 195Matches NCERT’s answer

1. b = 3 and c = 4 are given, and the foci are on the x-axis, so the major axis is along x.
2. a² = b² + c² = 9 + 16 = 25.
3. Write x²/25 + y²/9 = 1.

Answerx²/25 + y²/9 = 1

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Open NCERT p. 195Matches NCERT’s answer

1. The centre is (0, 0) and the major axis is on the y-axis, so the equation is x²/b² + y²/a² = 1, with a² > b².
2. Put in (3, 2): 9/b² + 4/a² = 1.
3. Put in (1, 6): 1/b² + 36/a² = 1.
4. Write u = 1/b² and v = 1/a². The equations become 9u + 4v = 1 and u + 36v = 1.
5. Multiply the second by 9: 9u + 324v = 9. Subtract the first: 320v = 8, so v = 1/40 and a² = 40.
6. Then u = 1 − 36/40 = 4/40 = 1/10, so b² = 10.
7. a² = 40 is bigger than b² = 10, so the major axis really is on the y-axis.
8. The equation is x²/10 + y²/40 = 1.

Answerx²/10 + y²/40 = 1

Watch this explained “Two places, one curve”, 13:13 into Putting the centre at the origin, and reading the axes off the equation

Open NCERT p. 195Matches NCERT’s answer

1. The major axis is on the x-axis (centre at the origin), so the equation is x²/a² + y²/b² = 1, with a² > b².
2. Put in (4, 3): 16/a² + 9/b² = 1.
3. Put in (6, 2): 36/a² + 4/b² = 1.
4. Write u = 1/a² and v = 1/b². The equations become 16u + 9v = 1 and 36u + 4v = 1.
5. Multiply the first by 4: 64u + 36v = 4. Multiply the second by 9: 324u + 36v = 9. Subtract: 260u = 5, so u = 1/52 and a² = 52.
6. Then 9v = 1 − 16/52 = 36/52, so v = 4/52 = 1/13 and b² = 13.
7. a² = 52 is bigger than b² = 13, so the major axis really is on the x-axis.
8. The equation is x²/52 + y²/13 = 1.

Answerx²/52 + y²/13 = 1

Watch this explained “Two places, one curve”, 13:13 into Putting the centre at the origin, and reading the axes off the equation