Mathematics Multiple Choice Questions on “Conic Sections – Hyperbola”.
1. A hyperbola has ___________ vertices and ____________ foci.
a) two, one
b) one, one
c) one, two
d) two, two
Answer: d
Clarification: A hyperbola has two vertices lying on each end and two foci lying inside the hyperbola.
If P is a point on hyperbola and F1 and F2 are foci then |PF1-PF2| remains constant.
2. The center of hyperbola is the same as a vertex.
a) True
b) False
Answer: b
Clarification: No, center and vertex are different for hyperbola.
Hyperbola has one center and two vertices.
3. Find the coordinates of foci of hyperbola ((frac{x}{9})^2-(frac{y}{16})^2)=1.
a) (±5,0)
b) (±4,0)
c) (0,±5)
d) (0,±4)
Answer: a
Clarification: Comparing the equation with ((frac{x}{a})^2-(frac{y}{b})^2)=1, we get a=3 and b=4.
For hyperbola, c2=a2+b2=9+16=25 => c=5.
So, coordinates of foci are (±c,0) i.e. (±5,0).
4. Find the coordinates of foci of hyperbola ((frac{y}{16})^2-(frac{x}{9})^2)=1.
a) (±5,0)
b) (±4,0)
c) (0,±5)
d) (0,±4)
Answer: c
Clarification: Comparing the equation with ((frac{y}{a})^2-(frac{x}{b})^2)=1, we get a=4 and b=3.
For hyperbola, c2=a2+b2= 16+9=25 => c=5.
So, coordinates of foci are (0,±c) i.e. (0,±5).
5. What is eccentricity for ((frac{x}{9})^2-(frac{y}{16})^2)=1?
a) 2/5
b) 3/5
c) 15
d) 5/3
Answer: d
Clarification: Comparing the equation with ((frac{x}{a})^2-(frac{y}{b})^2)=1, we get a=3 and b=4.
For hyperbola, c2=a2+b2= 9+16=25 => c=5.
We know, for hyperbola c=a*e
So, e=c/a = 5/3.
6. What is transverse axis length for hyperbola ((frac{x}{9})^2-(frac{y}{16})^2)=1?
a) 5 units
b) 4 units
c) 8 units
d) 6 units
Answer: d
Clarification: Comparing the equation with ((frac{x}{a})^2-(frac{y}{b})^2)=1, we get a=3 and b=4.
Transverse axis length = 2a = 2*3 =6 units.
7. What is conjugate axis length for hyperbola ((frac{x}{9})^2-(frac{y}{16})^2)=1?
a) 5 units
b) 4 units
c) 8 units
d) 10 units
Answer: c
Clarification: Comparing the equation with ((frac{x}{a})^2-(frac{y}{b})^2)=1, we get a=3 and b=4.
Conjugate axis length = 2b = 2*4 =8 units.
8. What is length of latus rectum for hyperbola ((frac{x}{9})^2-(frac{y}{16})^2)=1?
a) 25/2
b) 32/3
c) 5/32
d) 8/5
Answer: b
Clarification: Comparing the equation with ((frac{x}{a})^2-(frac{y}{b})^2)=1, we get a=3 and b=4.
We know, length of latus rectum = 2b2/a.
So, length of latus rectum of given hyperbola = 2*42/3 = 32/3.
9. What is equation of latus rectums of hyperbola ((frac{x}{9})^2-(frac{y}{16})^2)=1?
a) x=±5
b) y=±5
c) x=±2
d) y=±2
Answer: a
Clarification: Comparing the equation with ((frac{x}{a})^2-(frac{y}{b})^2)=1, we get a=3 and b=4.
For hyperbola, c2=a2+b2= 9+16=25 => c=5.
Equation of latus rectum x=±c i.e. x= ±5.
10. If length of transverse axis is 8 and conjugate axis is 10 and transverse axis is along x-axis then find the equation of hyperbola.
a) ((frac{x}{4})^2-(frac{y}{5})^2)=1
b) ((frac{x}{5})^2-(frac{y}{4})^2)=1
c) ((frac{x}{10})^2-(frac{y}{8})^2)=1
d) ((frac{x}{8})^2-(frac{y}{10})^2)=1
Answer: a
Clarification: Given, 2a=8 => a=4 and 2b=10 => b=5.
Equation of hyperbola with transverse axis along x-axis is ((frac{x}{4})^2-(frac{y}{5})^2)=1.
So, equation of given hyperbola is ((frac{x}{4})^2-(frac{y}{5})^2)=1.
11. If foci of a hyperbola are (0, ±5) and length of semi transverse axis is 3 units, then find the equation of hyperbola.
a) ((frac{x}{4})^2-(frac{y}{3})^2)=1
b) ((frac{x}{3})^2-(frac{y}{4})^2)=1
c) ((frac{x}{10})^2+(frac{y}{8})^2)=1
d) ((frac{x}{8})^2-(frac{y}{6})^2)=1
Answer: a
Clarification: Given, a=3 and c=5 => b2=c2-a2 = 52-32=42 => b=4.
Equation of hyperbola with transverse axis along y-axis is ((frac{y}{a})^2-(frac{x}{b})^2)=1.
So, equation of given hyperbola is ((frac{y}{3})^2-(frac{x}{4})^2)=1.
12. A hyperbola in which length of transverse and conjugate axis are equal is called _________ hyperbola.
a) isosceles
b) equilateral
c) bilateral
d) right
Answer: b
Clarification: A hyperbola in which length of transverse and conjugate axis is equal is called equilateral hyperbola. In this type of hyperbola, a=b i.e. 2a=2b or length of transverse and conjugate axis are equal.