Commerce Exam > Commerce Notes > Mathematics (Maths) Class 11 > RD Sharma Class 11 Solutions Chapter - Hyperbola
RD Sharma Class 11 Solutions Chapter - Hyperbola | Mathematics (Maths) Class 11 - Commerce PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
27. Hyperbola
Exercise 27.1
1. Question
The equation of the directrix of a hyperbola is x – y + 3 = 0. Its focus is (-1, 1) and eccentricity 3. Find the
equation of the hyperbola.
Answer
Given: Equation of directrix of a hyperbola is x – y + 3 = 0. Focus of hyperbola is (-1, 1) and eccentricity (e)
= 3
To find: equation of the hyperbola
Let M be the point on directrix and P(x, y) be any point of the hyperbola
Formula used:
where e is an eccentricity, PM is perpendicular from any point P on hyperbola to the directrix
Therefore,
Squaring both sides:
{? (a – b)
2
= a
2
+ b
2
+ 2ab &
(a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ac}
? 2{x
2
+ 1 + 2x + y
2
+ 1 – 2y} = 9{x
2
+ y
2
+ 9 + 6x – 6y – 2xy}
? 2x
2
+ 2 + 4x + 2y
2
+ 2 – 4y = 9x
2
+ 9y
2
+ 81 + 54x – 54y – 18xy
? 2x
2
+ 4 + 4x + 2y
2
– 4y – 9x
2
- 9y
2
- 81 – 54x + 54y + 18xy = 0
? – 7x
2
- 7y
2
– 50x + 50y + 18xy – 77 = 0
? 7x
2
+ 7y
2
+ 50x – 50y – 18xy + 77 = 0
This is the required equation of hyperbola
Page 2
27. Hyperbola
Exercise 27.1
1. Question
The equation of the directrix of a hyperbola is x – y + 3 = 0. Its focus is (-1, 1) and eccentricity 3. Find the
equation of the hyperbola.
Answer
Given: Equation of directrix of a hyperbola is x – y + 3 = 0. Focus of hyperbola is (-1, 1) and eccentricity (e)
= 3
To find: equation of the hyperbola
Let M be the point on directrix and P(x, y) be any point of the hyperbola
Formula used:
where e is an eccentricity, PM is perpendicular from any point P on hyperbola to the directrix
Therefore,
Squaring both sides:
{? (a – b)
2
= a
2
+ b
2
+ 2ab &
(a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ac}
? 2{x
2
+ 1 + 2x + y
2
+ 1 – 2y} = 9{x
2
+ y
2
+ 9 + 6x – 6y – 2xy}
? 2x
2
+ 2 + 4x + 2y
2
+ 2 – 4y = 9x
2
+ 9y
2
+ 81 + 54x – 54y – 18xy
? 2x
2
+ 4 + 4x + 2y
2
– 4y – 9x
2
- 9y
2
- 81 – 54x + 54y + 18xy = 0
? – 7x
2
- 7y
2
– 50x + 50y + 18xy – 77 = 0
? 7x
2
+ 7y
2
+ 50x – 50y – 18xy + 77 = 0
This is the required equation of hyperbola
2 A. Question
Find the equation of the hyperbola whose
focus is (0, 3), directrix is x + y – 1 = 0 and eccentricity = 2
Answer
Given: Equation of directrix of a hyperbola is x + y – 1 = 0. Focus of hyperbola is (0, 3) and eccentricity (e) =
2
To find: equation of the hyperbola
Let M be the point on directrix and P(x, y) be any point of the hyperbola
Formula used:
where e is an eccentricity, PM is perpendicular from any point P on hyperbola to the directrix
Therefore,
Squaring both sides:
{? (a – b)
2
= a
2
+ b
2
+ 2ab &
(a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ac}
? 2{x
2
+ y
2
+ 9 – 6y} = 4{x
2
+ y
2
+ 1 – 2x – 2y + 2xy}
Page 3
27. Hyperbola
Exercise 27.1
1. Question
The equation of the directrix of a hyperbola is x – y + 3 = 0. Its focus is (-1, 1) and eccentricity 3. Find the
equation of the hyperbola.
Answer
Given: Equation of directrix of a hyperbola is x – y + 3 = 0. Focus of hyperbola is (-1, 1) and eccentricity (e)
= 3
To find: equation of the hyperbola
Let M be the point on directrix and P(x, y) be any point of the hyperbola
Formula used:
where e is an eccentricity, PM is perpendicular from any point P on hyperbola to the directrix
Therefore,
Squaring both sides:
{? (a – b)
2
= a
2
+ b
2
+ 2ab &
(a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ac}
? 2{x
2
+ 1 + 2x + y
2
+ 1 – 2y} = 9{x
2
+ y
2
+ 9 + 6x – 6y – 2xy}
? 2x
2
+ 2 + 4x + 2y
2
+ 2 – 4y = 9x
2
+ 9y
2
+ 81 + 54x – 54y – 18xy
? 2x
2
+ 4 + 4x + 2y
2
– 4y – 9x
2
- 9y
2
- 81 – 54x + 54y + 18xy = 0
? – 7x
2
- 7y
2
– 50x + 50y + 18xy – 77 = 0
? 7x
2
+ 7y
2
+ 50x – 50y – 18xy + 77 = 0
This is the required equation of hyperbola
2 A. Question
Find the equation of the hyperbola whose
focus is (0, 3), directrix is x + y – 1 = 0 and eccentricity = 2
Answer
Given: Equation of directrix of a hyperbola is x + y – 1 = 0. Focus of hyperbola is (0, 3) and eccentricity (e) =
2
To find: equation of the hyperbola
Let M be the point on directrix and P(x, y) be any point of the hyperbola
Formula used:
where e is an eccentricity, PM is perpendicular from any point P on hyperbola to the directrix
Therefore,
Squaring both sides:
{? (a – b)
2
= a
2
+ b
2
+ 2ab &
(a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ac}
? 2{x
2
+ y
2
+ 9 – 6y} = 4{x
2
+ y
2
+ 1 – 2x – 2y + 2xy}
? 2x
2
+ 2y
2
+ 18 – 12y = 4x
2
+ 4y
2
+ 4 – 8x – 8y + 8xy
? 2x
2
+ 2y
2
+ 18 – 12y – 4x
2
– 4y
2
– 4 – 8x + 8y – 8xy = 0
? – 2x
2
– 2y
2
– 8x – 4y – 8xy + 14 = 0
? -2(x
2
+ y
2
– 4x + 2y + 4xy – 7) = 0
? x
2
+ y
2
– 4x + 2y + 4xy – 7 = 0
This is the required equation of hyperbola
2 B. Question
Find the equation of the hyperbola whose
focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity = 2
Answer
Given: Equation of directrix of a hyperbola is 3x + 4y + 8 = 0. Focus of hyperbola is (1, 1) and eccentricity
(e) = 2
To find: equation of hyperbola
Let M be the point on directrix and P(x, y) be any point of hyperbola
Formula used:
where e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix
Therefore,
Squaring both sides:
Page 4
27. Hyperbola
Exercise 27.1
1. Question
The equation of the directrix of a hyperbola is x – y + 3 = 0. Its focus is (-1, 1) and eccentricity 3. Find the
equation of the hyperbola.
Answer
Given: Equation of directrix of a hyperbola is x – y + 3 = 0. Focus of hyperbola is (-1, 1) and eccentricity (e)
= 3
To find: equation of the hyperbola
Let M be the point on directrix and P(x, y) be any point of the hyperbola
Formula used:
where e is an eccentricity, PM is perpendicular from any point P on hyperbola to the directrix
Therefore,
Squaring both sides:
{? (a – b)
2
= a
2
+ b
2
+ 2ab &
(a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ac}
? 2{x
2
+ 1 + 2x + y
2
+ 1 – 2y} = 9{x
2
+ y
2
+ 9 + 6x – 6y – 2xy}
? 2x
2
+ 2 + 4x + 2y
2
+ 2 – 4y = 9x
2
+ 9y
2
+ 81 + 54x – 54y – 18xy
? 2x
2
+ 4 + 4x + 2y
2
– 4y – 9x
2
- 9y
2
- 81 – 54x + 54y + 18xy = 0
? – 7x
2
- 7y
2
– 50x + 50y + 18xy – 77 = 0
? 7x
2
+ 7y
2
+ 50x – 50y – 18xy + 77 = 0
This is the required equation of hyperbola
2 A. Question
Find the equation of the hyperbola whose
focus is (0, 3), directrix is x + y – 1 = 0 and eccentricity = 2
Answer
Given: Equation of directrix of a hyperbola is x + y – 1 = 0. Focus of hyperbola is (0, 3) and eccentricity (e) =
2
To find: equation of the hyperbola
Let M be the point on directrix and P(x, y) be any point of the hyperbola
Formula used:
where e is an eccentricity, PM is perpendicular from any point P on hyperbola to the directrix
Therefore,
Squaring both sides:
{? (a – b)
2
= a
2
+ b
2
+ 2ab &
(a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ac}
? 2{x
2
+ y
2
+ 9 – 6y} = 4{x
2
+ y
2
+ 1 – 2x – 2y + 2xy}
? 2x
2
+ 2y
2
+ 18 – 12y = 4x
2
+ 4y
2
+ 4 – 8x – 8y + 8xy
? 2x
2
+ 2y
2
+ 18 – 12y – 4x
2
– 4y
2
– 4 – 8x + 8y – 8xy = 0
? – 2x
2
– 2y
2
– 8x – 4y – 8xy + 14 = 0
? -2(x
2
+ y
2
– 4x + 2y + 4xy – 7) = 0
? x
2
+ y
2
– 4x + 2y + 4xy – 7 = 0
This is the required equation of hyperbola
2 B. Question
Find the equation of the hyperbola whose
focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity = 2
Answer
Given: Equation of directrix of a hyperbola is 3x + 4y + 8 = 0. Focus of hyperbola is (1, 1) and eccentricity
(e) = 2
To find: equation of hyperbola
Let M be the point on directrix and P(x, y) be any point of hyperbola
Formula used:
where e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix
Therefore,
Squaring both sides:
{? (a – b)
2
= a
2
+ b
2
+ 2ab &
(a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ac}
? 25{x
2
+ 1 – 2x + y
2
+ 1 – 2y} = 4{9x
2
+ 16y
2
+ 64 + 24xy + 64y + 48x}
? 25x
2
+ 25 – 50x + 25y
2
+ 25 – 50y = 36x
2
+ 64y
2
+ 256 + 96xy + 256y + 192x
? 25x
2
+ 25 – 50x + 25y
2
+ 25 – 50y – 36x
2
– 64y
2
– 256 – 96xy – 256y – 192x = 0
? – 11x
2
– 39y
2
– 242x – 306y – 96xy – 206 = 0
? 11x
2
+ 39y
2
+ 242x + 306y + 96xy + 206 = 0
This is the required equation of hyperbola
2 C. Question
Find the equation of the hyperbola whose
focus is (1, 1) directrix is 2x + y = 1 and eccentricity =
Answer
Given: Equation of directrix of a hyperbola is 2x + y – 1 = 0. Focus of hyperbola is (1, 1) and eccentricity (e)
=
To find: equation of hyperbola
Let M be the point on directrix and P(x, y) be any point of hyperbola
Formula used:
where e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix
Therefore,
Page 5
27. Hyperbola
Exercise 27.1
1. Question
The equation of the directrix of a hyperbola is x – y + 3 = 0. Its focus is (-1, 1) and eccentricity 3. Find the
equation of the hyperbola.
Answer
Given: Equation of directrix of a hyperbola is x – y + 3 = 0. Focus of hyperbola is (-1, 1) and eccentricity (e)
= 3
To find: equation of the hyperbola
Let M be the point on directrix and P(x, y) be any point of the hyperbola
Formula used:
where e is an eccentricity, PM is perpendicular from any point P on hyperbola to the directrix
Therefore,
Squaring both sides:
{? (a – b)
2
= a
2
+ b
2
+ 2ab &
(a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ac}
? 2{x
2
+ 1 + 2x + y
2
+ 1 – 2y} = 9{x
2
+ y
2
+ 9 + 6x – 6y – 2xy}
? 2x
2
+ 2 + 4x + 2y
2
+ 2 – 4y = 9x
2
+ 9y
2
+ 81 + 54x – 54y – 18xy
? 2x
2
+ 4 + 4x + 2y
2
– 4y – 9x
2
- 9y
2
- 81 – 54x + 54y + 18xy = 0
? – 7x
2
- 7y
2
– 50x + 50y + 18xy – 77 = 0
? 7x
2
+ 7y
2
+ 50x – 50y – 18xy + 77 = 0
This is the required equation of hyperbola
2 A. Question
Find the equation of the hyperbola whose
focus is (0, 3), directrix is x + y – 1 = 0 and eccentricity = 2
Answer
Given: Equation of directrix of a hyperbola is x + y – 1 = 0. Focus of hyperbola is (0, 3) and eccentricity (e) =
2
To find: equation of the hyperbola
Let M be the point on directrix and P(x, y) be any point of the hyperbola
Formula used:
where e is an eccentricity, PM is perpendicular from any point P on hyperbola to the directrix
Therefore,
Squaring both sides:
{? (a – b)
2
= a
2
+ b
2
+ 2ab &
(a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ac}
? 2{x
2
+ y
2
+ 9 – 6y} = 4{x
2
+ y
2
+ 1 – 2x – 2y + 2xy}
? 2x
2
+ 2y
2
+ 18 – 12y = 4x
2
+ 4y
2
+ 4 – 8x – 8y + 8xy
? 2x
2
+ 2y
2
+ 18 – 12y – 4x
2
– 4y
2
– 4 – 8x + 8y – 8xy = 0
? – 2x
2
– 2y
2
– 8x – 4y – 8xy + 14 = 0
? -2(x
2
+ y
2
– 4x + 2y + 4xy – 7) = 0
? x
2
+ y
2
– 4x + 2y + 4xy – 7 = 0
This is the required equation of hyperbola
2 B. Question
Find the equation of the hyperbola whose
focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity = 2
Answer
Given: Equation of directrix of a hyperbola is 3x + 4y + 8 = 0. Focus of hyperbola is (1, 1) and eccentricity
(e) = 2
To find: equation of hyperbola
Let M be the point on directrix and P(x, y) be any point of hyperbola
Formula used:
where e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix
Therefore,
Squaring both sides:
{? (a – b)
2
= a
2
+ b
2
+ 2ab &
(a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ac}
? 25{x
2
+ 1 – 2x + y
2
+ 1 – 2y} = 4{9x
2
+ 16y
2
+ 64 + 24xy + 64y + 48x}
? 25x
2
+ 25 – 50x + 25y
2
+ 25 – 50y = 36x
2
+ 64y
2
+ 256 + 96xy + 256y + 192x
? 25x
2
+ 25 – 50x + 25y
2
+ 25 – 50y – 36x
2
– 64y
2
– 256 – 96xy – 256y – 192x = 0
? – 11x
2
– 39y
2
– 242x – 306y – 96xy – 206 = 0
? 11x
2
+ 39y
2
+ 242x + 306y + 96xy + 206 = 0
This is the required equation of hyperbola
2 C. Question
Find the equation of the hyperbola whose
focus is (1, 1) directrix is 2x + y = 1 and eccentricity =
Answer
Given: Equation of directrix of a hyperbola is 2x + y – 1 = 0. Focus of hyperbola is (1, 1) and eccentricity (e)
=
To find: equation of hyperbola
Let M be the point on directrix and P(x, y) be any point of hyperbola
Formula used:
where e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix
Therefore,
Squaring both sides:
{? (a – b)
2
= a
2
+ b
2
+ 2ab &
(a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2ab + 2bc + 2ac}
? 5{x
2
+ 1 – 2x + y
2
+ 1 – 2y} = 3{4x
2
+ y
2
+ 1 + 4xy – 2y – 4x}
? 5x
2
+ 5 – 10x + 5y
2
+ 5 – 10y = 12x
2
+ 3y
2
+ 3 + 12xy – 6y – 12x
? 5x
2
+ 5 – 10x + 5y
2
+ 5 – 10y – 12x
2
– 3y
2
– 3 – 12xy + 6y + 12x = 0
? – 7x
2
+ 2y
2
+ 2x – 4y – 12xy + 7 = 0
? 7x
2
– 2y
2
– 2x + 4y + 12xy – 7 = 0
This is the required equation of hyperbola.
2 D. Question
Find the equation of the hyperbola whose
focus is (2, -1), directrix is 2x + 3y = 1 and eccentricity = 2
Answer
Given: Equation of directrix of a hyperbola is 2x + 3y – 1 = 0. Focus of hyperbola is (2, -1) and eccentricity
(e) = 2
To find: equation of hyperbola
Let M be the point on directrix and P(x, y) be any point of hyperbola
Read More
|
75 videos|238 docs|91 tests
|
About this Document
|
Nov 23, 2024 Last updated |
Document Description: RD Sharma Class 11 Solutions Chapter - Hyperbola for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation.
The notes and questions for RD Sharma Class 11 Solutions Chapter - Hyperbola have been prepared according to the Commerce exam syllabus. Information about RD Sharma Class 11 Solutions Chapter - Hyperbola covers topics
like and RD Sharma Class 11 Solutions Chapter - Hyperbola Example, for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Class 11 Solutions Chapter - Hyperbola.
Introduction of RD Sharma Class 11 Solutions Chapter - Hyperbola in English is available as part of our Mathematics (Maths) Class 11
for Commerce & RD Sharma Class 11 Solutions Chapter - Hyperbola in Hindi for Mathematics (Maths) Class 11 course.
Download more important topics related with notes, lectures and mock test series for Commerce
Exam by signing up for free. Commerce: RD Sharma Class 11 Solutions Chapter - Hyperbola | Mathematics (Maths) Class 11 - Commerce
Description
Full syllabus notes, lecture & questions for RD Sharma Class 11 Solutions Chapter - Hyperbola | Mathematics (Maths) Class 11 - Commerce - Commerce | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 11 | Best notes, free PDF download
Information about RD Sharma Class 11 Solutions Chapter - Hyperbola
In this doc you can find the meaning of RD Sharma Class 11 Solutions Chapter - Hyperbola defined & explained in the simplest way possible. Besides explaining types of
RD Sharma Class 11 Solutions Chapter - Hyperbola theory, EduRev gives you an ample number of questions to practice RD Sharma Class 11 Solutions Chapter - Hyperbola tests, examples and also practice Commerce
tests
|
Explore Courses for Commerce exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
practice quizzes
,Semester Notes
,RD Sharma Class 11 Solutions Chapter - Hyperbola | Mathematics (Maths) Class 11 - Commerce
,Previous Year Questions with Solutions
,Exam
,mock tests for examination
,Objective type Questions
,ppt
,Sample Paper
,video lectures
,past year papers
,RD Sharma Class 11 Solutions Chapter - Hyperbola | Mathematics (Maths) Class 11 - Commerce
,Viva Questions
,RD Sharma Class 11 Solutions Chapter - Hyperbola | Mathematics (Maths) Class 11 - Commerce
,Important questions
,Summary
,shortcuts and tricks
,MCQs
,Free
,Extra Questions
,study material
;
Additional Information about RD Sharma Class 11 Solutions Chapter - Hyperbola for Commerce Preparation
RD Sharma Class 11 Solutions Chapter - Hyperbola Free PDF Download
The RD Sharma Class 11 Solutions Chapter - Hyperbola is an invaluable resource that delves deep into the core of the Commerce exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the RD Sharma Class 11 Solutions Chapter - Hyperbola now and kickstart your journey towards success in the Commerce exam.
Importance of RD Sharma Class 11 Solutions Chapter - Hyperbola
The importance of RD Sharma Class 11 Solutions Chapter - Hyperbola cannot be overstated, especially for Commerce aspirants.
This document holds the key to success in the Commerce exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
RD Sharma Class 11 Solutions Chapter - Hyperbola Notes
RD Sharma Class 11 Solutions Chapter - Hyperbola Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RD Sharma Class 11 Solutions Chapter - Hyperbola.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, RD Sharma Class 11 Solutions Chapter - Hyperbola Notes on EduRev are your ultimate resource for success.
RD Sharma Class 11 Solutions Chapter - Hyperbola Commerce Questions
The "RD Sharma Class 11 Solutions Chapter - Hyperbola Commerce Questions" guide is a valuable resource for all aspiring students preparing for the
Commerce exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study RD Sharma Class 11 Solutions Chapter - Hyperbola on the App
Students of Commerce can study RD Sharma Class 11 Solutions Chapter - Hyperbola alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RD Sharma Class 11 Solutions Chapter - Hyperbola,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of RD Sharma Class 11 Solutions Chapter - Hyperbola is prepared as per the latest Commerce syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup