Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Find the equation of the hyperbola whose asymptotes are 2x – y = 3 and 3x + y – 7 = 0 and which passes through the point (1, 1). Find the equation of the corresponding conjugate hyperbola.

Option: 1

 (2x – y – 3) (3x + y – 7) = 6


Option: 2

 (2x – y – 3) (3x + y – 7) = -6


Option: 3

(2x – y – 3) (3x + y – 7) = 0


Option: 4

 none of these


Answers (1)

As we learned

Conjugate Hyperbola -

\frac {y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}= 1

- wherein

 

 

The equation of the hyperbola differs from the equation of the asymptotes by a constant

The equation of the hyperbola with asymptotes 3x + y – 7 = 0 and 2x – y = 3 is

(3x + y – 7) (2x – y – 3) + k = 0

It passes through (1, 1)

k = –6

Hence the equation of the hyperbola is (2x – y – 3) (3x + y – 7) = 6

The equation of the conjugate hyperbola is (3x + y – 7) (2x – y – 3) – k = 0

i.e., (3x + y – 7) (2x – y – 3) + 6 = 0

Posted by

Kshitij

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions