Engineering Drawing Questions and Answers – Construction of Hyperbola – 2

This set of Engineering Drawing Multiple Choice Questions & Answers (MCQs) focuses on “Construction of Hyperbola – 2”.

1. Hyperbolic curves are used for which of the following designs?
a) Light reflectors
b) Sound reflectors
c) Cooling towers
d) Arches
View Answer

Answer: c
Explanation: The engineering curves play a major role in designing. Hyperbolic curves are one of the most important engineering curves which are used in the design of cooling towers and water channels.

2. In which of the following hyperbolic curves are not used in the design phase?
a) Cooling towers
b) Water channels
c) Stuffing box
d) Mirrors in long-distance telescopes
View Answer

Answer: c
Explanation: The engineering curves play a major role in designing. Hyperbolic curves are used in the design of the water channels, cooling towers, mirror used in the long-distance telescope. Whereas stuffing box design requires the elliptical curves.

3. Which of the following described cut sections gives the hyperbolic curve from the cone?
a) A section parallel to the cone axis
b) A section parallel to the base of the cone
c) A section parallel to one of the generators
d) Section perpendicular to the cone axis
View Answer

Answer: a
Explanation: Hyperbolic curves are the conic sections which are formed when the section parallel or inclined to the axis of the cone cut it on one side. When the section is parallel to the axis, then rectangular hyperbola is formed.
advertisement
advertisement

4. Rectangular hyperbola is formed when the section plane is parallel to the axis.
a) True
b) False
View Answer

Answer: a
Explanation: When a section plane which is parallel or inclined to the cone axis, a hyperbola is formed. When the cut section is parallel to the cone axis then a rectangular hyperbola is formed.

5. Which of the following is the value of eccentricity which suits the hyperbola?
a) 1
b) 1/2
c) 3/2
d) 1/4
View Answer

Answer: c
Explanation: The eccentricity is defined as the ratio of the distance of the point from the focus to the distance of the point from the directrix. For the hyperbola the value of eccentricity e>1, hence 3/2 is the eccentricity value of a hyperbola.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

6. Which of the following is the standard hyperbolical curve equation?
a) X2 + Y2 = a2
b) X2 + Y2 = a
c) X2/a2 + Y2b2 = 1
d) X2/a2 – Y2/b2 = 1
View Answer

Answer: d
Explanation: Hyperbola is one of the most important curves of the engineering drawing. The general standard form of the hyperbola is X2/a2 – Y2/b2 = 1, where a, b are constants. It is the conical form of a hyperbola. Any type of hyperbolic equations can be modified and transformed into this standard form.

7. Which of the following equation is true for the hyperbola, where P is on the hyperbola curve, F1, F2 are the two fixed points, the foci, and a is the distance from the vertices of the hyperbola to its center.
a) PF1 – PF2 = a
b) ||PF2 |-|PF1|| = 2a
c) |PF1 – PF2 | = 2a
d) |PF2 – PF1 | = 2a
View Answer

Answer: b
Explanation: The hyperbola is the curve with the locus points having an absolute difference of the distance from the focus that is ||PF2 |-|PF1|| is a constant and is equal to the distance between the vertices that is 2a.
advertisement

8. The rectangular hyperbola is the set of points whose product of the distance from the fixed lines perpendicular to each other is constant. What are those fixed lines called?
a) Directrix
b) Latus rectum
c) Asymptotes
d) Semi-latus rectum
View Answer

Answer: d
Explanation: Rectangular hyperbola is the curve which is the collection of points whose distances from the two fixed lines called asymptotes is constant. It has a constant eccentricity of 21/2 and the angle between the asymptotes is 90°.

9. Which of the following is the eccentricity of a rectangular hyperbola?
a) 1
b) 1/2
c) 3/2
d) √2
View Answer

Answer: d
Explanation: The eccentricity is defined as the ratio of the distance of the point from the focus to the distance of the point from the directrix. For the hyperbola the value of eccentricity e>1, it is defined as e = (1 + b2/a2)1/2, whereas in rectangular hyperbola a=b hence the value of eccentricity is constant and equal to √2.
advertisement

10. Find the eccentricity of the hyperbola of equation X2/32 – Y2/42 = 1?
a) 3/2
b) 3/4
c) 4/3
d) 5/4
View Answer

Answer: d
Explanation: The general standard form of the hyperbola is X2/a2 – Y2/b2 = 1, where a, b are constants. For the hyperbola the value of eccentricity e>1, it is defined as e = (1 + b2/a2)1/2, where b=4 and a=3 hence e = (1 + 42/32)1/2 = (52/42)1/2. Hence e = 5/4.

11. As per Boyle’s law the product of pressure and volume, P*V = constant. In water channels, we use this equation which results in ______
a) Parabolic curve
b) Rectangular hyperbolic curve
c) Elliptical curve
d) Straight line
View Answer

Answer: b
Explanation: As per the given equation P*V = constant results in rectangular hyperbola, which is the collection of points whose distances from the two fixed lines called asymptotes is constant. Hence in the design of water channels, we use the rectangular hyperbola.

12. For the equation of hyperbola XY = 3, what will be the eccentricity?
a) √2
b) √3
c) 1
d) 1/2
View Answer

Answer: a
Explanation: The given hyperbolic equation belongs to a rectangular hyperbola, whose eccentricity is constant that is 20.5. Hence for the hyperbola XY = 3 is √2.

13. The equation for the asymptotes of the hyperbola of equation X2/a2 – Y2/b2 = 1 is ______
a) Y = -bX/a
b) Y = bX/a
c) Y = ±bX/a
d) Y = aX/b
View Answer

Answer: c
Explanation: Hyperbola of equation X2/a2 – Y2/b2 = 1, when solved for Y we get ±b(X2-a2)0.5/a. Asymptotes are the tangents at infinite point of contact and they intersect at the center of the parabola, hence the equation of pair of asymptotes is Y = ±bX/a.

14. Find the eccentricity of the hyperbola of equation X2/4 – Y2/5 =1?
a) 3/2
b) 3/4
c) 4/3
d) 5/4
View Answer

Answer: a
Explanation: The general standard form of the hyperbola is X2/a2 – Y2/b2 = 1, where a, b are constants. For the hyperbola the value of eccentricity is defined as e = (1 + b2/a2)1/2, where b = 50.5 and a = 2 hence e = (1 + 5/22)1/2 = (9/22)1/2. Hence e = 3/2.

15. Find the conjugate hyperbola equation for the hyperbola of equation X2/a2 – Y2/b2 = 1?
a) X2/b2 – Y2/a2 = -1
b) X2/a2 – Y2/b2 = -1
c) X2/a2 – Y2/b2 = 0
d) X2/b2 – Y2/a2 = 0
View Answer

Answer: a
Explanation: Conjugate hyperbola is formed by exchange of x and y to obtain the equation. The given hyperbola equation is X2/a2 – Y2/b2 = 1, hence we get Y2/a2 – X2/b2 = 1, by transforming we get X2/b2 – Y2/a2 = -1.

Sanfoundry Global Education & Learning Series – Engineering Drawing.

To practice all areas of Engineering Drawing, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.