Find the coordinates of the foci, vertices, eccentricity, and latus rectum of the hyperbola x^2/16-y^2/9=1.
Hint. Use c^2=a^2+b^2 for a hyperbola, not a^2-b^2.
a^2=16, b^2=9, so a=4, b=3. c=sqrt(16+9)=5.
✦ Working through each part gives: Foci (+-5,0); vertices (+-4,0); eccentricity e=5/4; latus rectum=2(9)/4=9/2.
Find the coordinates of the foci, vertices, eccentricity, and latus rectum of the hyperbola y^2/9-x^2/27=1.
Hint. The positive term (y^2 here) marks the transverse axis, so it's vertical.
a^2=9, b^2=27, so a=3, b=3sqrt(3). c=sqrt(9+27)=6.
✦ Working through each part gives: Foci (0,+-6); vertices (0,+-3); eccentricity e=6/3=2; latus rectum=2(27)/3=18.
Find the coordinates of the foci, vertices, eccentricity, and latus rectum of the hyperbola 9y^2-4x^2=36.
Hint. Divide through by 36 to reach standard form first.
Dividing by 36: y^2/4-x^2/9=1. a^2=4, b^2=9, so a=2, b=3. c=sqrt(4+9)=sqrt(13).
✦ Working through each part gives: Foci (0,+-sqrt(13)); vertices (0,+-2); eccentricity e=sqrt(13)/2; latus rectum=2(9)/2=9.
Find the coordinates of the foci, vertices, eccentricity, and latus rectum of the hyperbola 16x^2-9y^2=576.
Hint. Divide through by 576 to reach standard form first.
Dividing by 576: x^2/36-y^2/64=1. a^2=36, b^2=64, so a=6, b=8. c=sqrt(36+64)=10.
✦ Working through each part gives: Foci (+-10,0); vertices (+-6,0); eccentricity e=10/6=5/3; latus rectum=2(64)/6=64/3.
Find the coordinates of the foci, vertices, eccentricity, and latus rectum of the hyperbola 5y^2-9x^2=36.
Hint. Divide through by 36 to reach standard form; the denominators will not be whole numbers here.
Dividing by 36: y^2/(36/5)-x^2/4=1. a^2=36/5, b^2=4, so a=6/sqrt(5)=6sqrt(5)/5. c^2=a^2+b^2=36/5+4=56/5, so c=sqrt(56/5)=2sqrt(70)/5.
✦ Working through each part gives: Foci (0,+-2sqrt(70)/5); vertices (0,+-6sqrt(5)/5); eccentricity e=sqrt(14)/3; latus rectum=2(4)/(6/sqrt(5))=4sqrt(5)/3.
Find the coordinates of the foci, vertices, eccentricity, and latus rectum of the hyperbola 49y^2-16x^2=784.
Hint. Divide through by 784 to reach standard form first.
Dividing by 784: y^2/16-x^2/49=1. a^2=16, b^2=49, so a=4, b=7. c=sqrt(16+49)=sqrt(65).
✦ Working through each part gives: Foci (0,+-sqrt(65)); vertices (0,+-4); eccentricity e=sqrt(65)/4; latus rectum=2(49)/4=49/2.
Find the equation of the hyperbola satisfying: vertices (+-2,0), foci (+-3,0).
Hint. Vertices give a, foci give c; use b^2=c^2-a^2.
Since the vertices are (+-2,0), a=2. Since the foci are (+-3,0), c=3. For a hyperbola, b^2=c^2-a^2 (not a^2-c^2, since c is always the larger of the two): b^2=9-4=5. As both lie on the x-axis, the transverse axis is along the x-axis.
✦ Working through each part gives: x^2/4-y^2/5=1.
Find the equation of the hyperbola satisfying: vertices (0,+-5), foci (0,+-8).
Hint. Vertices on the y-axis mean the transverse axis is vertical.
Since the vertices are (0,+-5), a=5. Since the foci are (0,+-8), c=8. Using b^2=c^2-a^2: b^2=64-25=39. As both the vertices and foci lie on the y-axis, the transverse axis is vertical, so the y^2 term is positive.
✦ Working through each part gives: y^2/25-x^2/39=1.
Find the equation of the hyperbola satisfying: vertices (0,+-3), foci (0,+-5).
Hint. Vertices on the y-axis mean the transverse axis is vertical.
Since the vertices are (0,+-3), a=3. Since the foci are (0,+-5), c=5. Using b^2=c^2-a^2: b^2=25-9=16. As both the vertices and foci lie on the y-axis, the transverse axis is vertical, so the y^2 term is positive.
✦ Working through each part gives: y^2/9-x^2/16=1.
Find the equation of the hyperbola satisfying: foci (+-5,0), transverse axis of length 8.
Hint. The transverse axis length is 2a, not a itself.
c=5, 2a=8 so a=4, giving b^2=25-16=9.
✦ Working through each part gives: x^2/16-y^2/9=1.
Find the equation of the hyperbola satisfying: foci (0,+-13), conjugate axis of length 24.
Hint. The conjugate axis length is 2b, not b itself. Foci on the y-axis mean the transverse axis is vertical.
c=13, 2b=24 so b=12, giving a^2=c^2-b^2=169-144=25.
✦ Working through each part gives: y^2/25-x^2/144=1.
Find the equation of the hyperbola satisfying: foci (+-3sqrt(5),0), latus rectum of length 8.
Hint. Write both conditions as equations in a and b, using c^2=a^2+b^2 to combine them.
c=3sqrt(5), so c^2=45. Latus rectum 2b^2/a=8 gives b^2=4a. Substituting into c^2=a^2+b^2: 45=a^2+4a, i.e. a^2+4a-45=0, which factors as (a+9)(a-5)=0. Since a>0, a=5, giving b^2=4(5)=20.
✦ Working through each part gives: x^2/25-y^2/20=1.
Find the equation of the hyperbola satisfying: foci (+-4,0), latus rectum of length 12.
Hint. Write both conditions as equations in a and b, using c^2=a^2+b^2 to combine them.
c=4, so c^2=16. Latus rectum 2b^2/a=12 gives b^2=6a. Substituting into c^2=a^2+b^2: 16=a^2+6a, i.e. a^2+6a-16=0, which factors as (a+8)(a-2)=0. Since a>0, a=2, giving b^2=6(2)=12.
✦ Working through each part gives: x^2/4-y^2/12=1.
Find the equation of the hyperbola satisfying: vertices (+-7,0), e=4/3.
Hint. Use e=c/a to find c from a, then b^2=c^2-a^2.
a=7. e=c/a=4/3 gives c=28/3. b^2=c^2-a^2=(28/3)^2-49=784/9-441/9=343/9.
✦ Working through each part gives: x^2/49-9y^2/343=1.
Find the equation of the hyperbola satisfying: foci (0,+-sqrt(10)), passing through (2,3).
Hint. Foci on the y-axis give the form y^2/a^2-x^2/b^2=1 with b^2=c^2-a^2=10-a^2. Substitute the given point and solve for a^2.
c^2=10. With b^2=10-a^2, substitute (2,3) into y^2/a^2-x^2/b^2=1: 9/a^2-4/(10-a^2)=1. Let A=a^2: 9(10-A)-4A=A(10-A), i.e. 90-13A=10A-A^2, giving A^2-23A+90=0. This factors using the quadratic formula: A=(23+-sqrt(529-360))/2=(23+-13)/2, so A=18 or A=5. Since b^2=10-A must be positive, A=18 is rejected (it gives b^2=-8), leaving A=5, so a^2=5 and b^2=10-5=5.
✦ Working through each part gives: y^2/5-x^2/5=1.
Where students slip. Taking both roots of the quadratic as valid answers without checking that b^2=10-a^2 stays positive lets through an impossible negative b^2.
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.
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