Exercise 10.3 — Conic Sections

1. 10.3.13 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   Find the coordinates of the foci, vertices, length of major axis, minor axis, eccentricity, and latus rectum of the ellipse x^2/36+y^2/16=1.

   Hint. Compare denominators: the larger one marks the major axis. Then use c^2=a^2-b^2.

   Since 36>16, the major axis is along the x-axis with a^2=36, b^2=16, so a=6, b=4. c=sqrt(36-16)=sqrt(20)=2sqrt(5).

   ✦ Working through each part gives: Foci (+-2sqrt(5),0); vertices (+-6,0); major axis=12, minor axis=8; eccentricity e=2sqrt(5)/6=sqrt(5)/3; latus rectum=2(16)/6=16/3.

2. 10.3.23 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   Find the coordinates of the foci, vertices, length of major axis, minor axis, eccentricity, and latus rectum of the ellipse x^2/4+y^2/25=1.

   Hint. Compare denominators: 25 is larger, so the major axis is along the y-axis this time.

   Since 25>4, the major axis is along the y-axis with a^2=25, b^2=4, so a=5, b=2. c=sqrt(25-4)=sqrt(21).

   ✦ Working through each part gives: Foci (0,+-sqrt(21)); vertices (0,+-5); major axis=10, minor axis=4; eccentricity e=sqrt(21)/5; latus rectum=2(4)/5=8/5.

3. 10.3.33 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   Find the coordinates of the foci, vertices, length of major axis, minor axis, eccentricity, and latus rectum of the ellipse x^2/16+y^2/9=1.

   Hint. Compare denominators: 16 is larger, major axis along the x-axis.

   Since 16>9, the major axis is along the x-axis with a^2=16, b^2=9, so a=4, b=3. c=sqrt(16-9)=sqrt(7).

   ✦ Working through each part gives: Foci (+-sqrt(7),0); vertices (+-4,0); major axis=8, minor axis=6; eccentricity e=sqrt(7)/4; latus rectum=2(9)/4=9/2.

4. 10.3.43 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   Find the coordinates of the foci, vertices, length of major axis, minor axis, eccentricity, and latus rectum of the ellipse x^2/25+y^2/100=1.

   Hint. Compare denominators: 100 is larger, major axis along the y-axis.

   Since 100>25, the major axis is along the y-axis with a^2=100, b^2=25, so a=10, b=5. c=sqrt(100-25)=sqrt(75)=5sqrt(3).

   ✦ Working through each part gives: Foci (0,+-5sqrt(3)); vertices (0,+-10); major axis=20, minor axis=10; eccentricity e=5sqrt(3)/10=sqrt(3)/2; latus rectum=2(25)/10=5.

5. 10.3.53 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   Find the coordinates of the foci, vertices, length of major axis, minor axis, eccentricity, and latus rectum of the ellipse x^2/49+y^2/36=1.

   Hint. Compare denominators: 49 is larger, major axis along the x-axis.

   Since 49>36, the major axis is along the x-axis with a^2=49, b^2=36, so a=7, b=6. c=sqrt(49-36)=sqrt(13).

   ✦ Working through each part gives: Foci (+-sqrt(13),0); vertices (+-7,0); major axis=14, minor axis=12; eccentricity e=sqrt(13)/7; latus rectum=2(36)/7=72/7.

6. 10.3.63 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   Find the coordinates of the foci, vertices, length of major axis, minor axis, eccentricity, and latus rectum of the ellipse x^2/100+y^2/400=1.

   Hint. Compare denominators: 400 is larger, major axis along the y-axis.

   Since 400>100, the major axis is along the y-axis with a^2=400, b^2=100, so a=20, b=10. c=sqrt(400-100)=sqrt(300)=10sqrt(3).

   ✦ Working through each part gives: Foci (0,+-10sqrt(3)); vertices (0,+-20); major axis=40, minor axis=20; eccentricity e=10sqrt(3)/20=sqrt(3)/2; latus rectum=2(100)/20=10.

7. 10.3.74 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   Find the coordinates of the foci, vertices, length of major axis, minor axis, eccentricity, and latus rectum of the ellipse 36x^2+4y^2=144.

   Hint. First divide through by 144 to get the standard form 1 on the right, then compare denominators.

   Dividing by 144: x^2/4+y^2/36=1. Since 36>4, the major axis is along the y-axis with a^2=36, b^2=4, so a=6, b=2. c=sqrt(36-4)=sqrt(32)=4sqrt(2).

   ✦ Working through each part gives: Foci (0,+-4sqrt(2)); vertices (0,+-6); major axis=12, minor axis=4; eccentricity e=4sqrt(2)/6=2sqrt(2)/3; latus rectum=2(4)/6=4/3.

8. 10.3.84 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   Find the coordinates of the foci, vertices, length of major axis, minor axis, eccentricity, and latus rectum of the ellipse 16x^2+y^2=16.

   Hint. Divide through by 16 to reach standard form.

   Dividing by 16: x^2/1+y^2/16=1. Since 16>1, the major axis is along the y-axis with a^2=16, b^2=1, so a=4, b=1. c=sqrt(16-1)=sqrt(15).

   ✦ Working through each part gives: Foci (0,+-sqrt(15)); vertices (0,+-4); major axis=8, minor axis=2; eccentricity e=sqrt(15)/4; latus rectum=2(1)/4=1/2.

9. 10.3.94 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

   Find the coordinates of the foci, vertices, length of major axis, minor axis, eccentricity, and latus rectum of the ellipse 4x^2+9y^2=36.

   Hint. Divide through by 36 to reach standard form.

   Dividing by 36: x^2/9+y^2/4=1. Since 9>4, the major axis is along the x-axis with a^2=9, b^2=4, so a=3, b=2. c=sqrt(9-4)=sqrt(5).

   ✦ Working through each part gives: Foci (+-sqrt(5),0); vertices (+-3,0); major axis=6, minor axis=4; eccentricity e=sqrt(5)/3; latus rectum=2(4)/3=8/3.

10. 10.3.103 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

    Find the equation of the ellipse that satisfies: vertices (+-5,0), foci (+-4,0).

    Hint. Vertices give a, foci give c; use b^2=a^2-c^2.

    a=5, c=4, so b^2=25-16=9. Since the vertices are on the x-axis, the major axis is along the x-axis.

    ✦ Working through each part gives: x^2/25+y^2/9=1.

11. 10.3.113 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

    Find the equation of the ellipse that satisfies: vertices (0,+-13), foci (0,+-5).

    Hint. Vertices give a, foci give c; use b^2=a^2-c^2. Vertices on the y-axis mean the major axis is vertical.

    a=13, c=5, so b^2=169-25=144. Since the vertices are on the y-axis, the major axis is along the y-axis.

    ✦ Working through each part gives: x^2/144+y^2/169=1.

12. 10.3.123 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

    Find the equation of the ellipse that satisfies: vertices (+-6,0), foci (+-4,0).

    Hint. Vertices give a, foci give c; use b^2=a^2-c^2.

    Since the vertices are (+-6,0), a=6. Since the foci are (+-4,0), c=4. Using b^2=a^2-c^2: b^2=36-16=20. As the vertices and foci both lie on the x-axis, the major axis is along the x-axis, so a^2 sits under x^2.

    ✦ Working through each part gives: x^2/36+y^2/20=1.

13. 10.3.133 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

    Find the equation of the ellipse that satisfies: ends of major axis (+-3,0), ends of minor axis (0,+-2).

    Hint. The ends of the major axis directly give a; the ends of the minor axis directly give b.

    a=3 (half the major axis), b=2 (half the minor axis). Since the major axis is on the x-axis, a^2 goes under x^2.

    ✦ Working through each part gives: x^2/9+y^2/4=1.

14. 10.3.143 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

    Find the equation of the ellipse that satisfies: ends of major axis (0,+-sqrt(5)), ends of minor axis (+-1,0).

    Hint. The major axis is vertical here, so a^2 goes under y^2.

    a=sqrt(5) (major axis along y), b=1 (minor axis along x).

    ✦ Working through each part gives: x^2/1+y^2/5=1.

15. 10.3.153 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

    Find the equation of the ellipse that satisfies: length of major axis 26, foci (+-5,0).

    Hint. The given length is 2a, not a itself.

    2a=26, so a=13. c=5, so b^2=169-25=144.

    ✦ Working through each part gives: x^2/169+y^2/144=1.

16. 10.3.163 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

    Find the equation of the ellipse that satisfies: length of minor axis 16, foci (0,+-6).

    Hint. The given length is 2b. Foci on the y-axis mean the major axis is vertical, so use a^2=b^2+c^2.

    2b=16, so b=8. c=6, and since the major axis is vertical, a^2=b^2+c^2=64+36=100.

    ✦ Working through each part gives: x^2/64+y^2/100=1.

17. 10.3.173 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

    Find the equation of the ellipse that satisfies: foci (+-3,0), a=4.

    Hint. c is given directly by the foci; use b^2=a^2-c^2.

    Since the foci are (+-3,0), c=3, and a is given directly as 4. Using b^2=a^2-c^2: b^2=16-9=7. Since the foci lie on the x-axis, the major axis is along the x-axis, so a^2 sits under x^2.

    ✦ Working through each part gives: x^2/16+y^2/7=1.

18. 10.3.183 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

    Find the equation of the ellipse that satisfies: b=3, c=4, centre at the origin, foci on the x-axis.

    Hint. Foci on the x-axis with both b and c given: use a^2=b^2+c^2.

    Here b=3 and c=4 are both given directly, so b^2=9 and c^2=16. Using a^2=b^2+c^2: a^2=9+16=25. Since the foci lie on the x-axis, the major axis is along the x-axis, so a^2=25 sits under x^2 and b^2=9 sits under y^2.

    ✦ Working through each part gives: x^2/25+y^2/9=1.

19. 10.3.195 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

    Find the equation of the ellipse with centre at (0,0), major axis on the y-axis, passing through the points (3,2) and (1,6).

    Hint. With the major axis on the y-axis, use x^2/b^2+y^2/a^2=1. Substitute both points to get a linear system in 1/a^2 and 1/b^2.

    The equation has the form x^2/b^2+y^2/a^2=1. Let u=1/b^2 and v=1/a^2. Substituting (3,2): 9u+4v=1. Substituting (1,6): u+36v=1. From the second equation, u=1-36v. Substituting into the first: 9(1-36v)+4v=1, so 9-324v+4v=1, giving -320v=-8, i.e. v=1/40. Then u=1-36/40=1/10. So b^2=1/u=10 and a^2=1/v=40.

    ✦ Working through each part gives: x^2/10+y^2/40=1.

20. 10.3.205 marksNCERT Class 11 Mathematics, Conic Sections, Reprint 2026-27

    Find the equation of the ellipse with major axis on the x-axis, passing through the points (4,3) and (6,2).

    Hint. With the major axis on the x-axis, use x^2/a^2+y^2/b^2=1. Substitute both points to get a linear system in 1/a^2 and 1/b^2.

    The equation has the form x^2/a^2+y^2/b^2=1. Let u=1/a^2 and v=1/b^2. Substituting (4,3): 16u+9v=1. Substituting (6,2): 36u+4v=1. Multiplying the first by 4 and the second by 9: 64u+36v=4 and 324u+36v=9. Subtracting: 260u=5, so u=1/52. Substituting back into 16u+9v=1: 16/52+9v=1, so 9v=1-4/13=9/13, giving v=1/13. So a^2=1/u=52 and b^2=1/v=13.

    ✦ Working through each part gives: x^2/52+y^2/13=1.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh110.pdf) — Exercise 10.1 (15 questions), Exercise 10.2 (12 questions), Exercise 10.3 (20 questions), Exercise 10.4 (15 questions), plus the chapter's Miscellaneous Exercise (8 questions), 70 questions total. Confirmed against the CBSE curriculum PDF that Conic Sections has no formative-only carve-out at all, unlike its neighbouring chapters Straight Lines and Introduction to Three-Dimensional Geometry — every phrase in its syllabus line is summative. Confirmed the book teaches the general circle equation only through completing the square (Example 3), never naming g, f, c as constants, and never mentions hyperbola asymptotes anywhere (zero hits for the word itself, and none of Exercise 10.4's 15 questions ask for one). Exercise 10.1 Q3's centre coordinates were cross-checked via a 300dpi page render after the raw PDF text extraction reversed their order. Several exercise answers (10.1 Q10/Q11, 10.4 Q15, Misc Q2/Q3/Q5/Q8) were cross-verified against known standard results for this exact problem set.. Questions are referenced from the NCERT textbook for identification.