wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The asymptotes of a hyperbola have center at the point (1,2) and are parallel to the lines 2x+3y=0 and 3x+2y=0. If the hyperbola passes through the point (5,3), then its equation is

A
(3x+2y8)(2x+3y7)156=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(2x+3y8)(3x+2y7)154=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(2x+3y7)(3x+2y+8)348=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(3x+2y8)(2x+3y+7)338=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B (2x+3y8)(3x+2y7)154=0
The equation of asymptotes parallel to the lines 2x+3y=0 and 3x+2y=0 are
2x+3y+λ=0(1)
3x+2y+μ=0(2)
It is given that these asmptotes pass through the center (1,2).
Substituting the point (1,2) in (1) and (2), we get
λ=8.μ=7

Thus, the combined equation of asymptotes is (2x+3y8)(3x+2y7)=0

Let the equation of the hyperbola be (2x+3y8)(3x+2y7)+ω=0
This hyperbola passes through (5,3)
(25+338)(35+237)+ω=0
ω=154

Hence, the equation of the hyperbola is (2x+3y8)(3x+2y7)154=0

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Hyperbola and Terminologies
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon