JEE MATHEMATICS MCQ: TRANSFORMATION OF AXES

1 Easy

1 of 1

Application +4 | -1

The pair of lines

\sqrt 3 {x^2} - 4xy + \sqrt 3 {y^2} = 0

are rotated about the origin by

\pi /6

in the anti-clockwise sense. The equation of the pair in the new position is

View answer and explanation

Correct option: A

\sqrt 3 {y^2} - xy = 0

Option A (

\sqrt 3 {y^2} - xy = 0

) is correct because after rotating the pair of lines by

\pi/6

in the anti-clockwise sense, the equation becomes

\sqrt 3 {y^2} - xy = 0

More Questions from TRANSFORMATION OF AXES

Practice All Chapter MCQs →

12 x^2+7 x y-12 y^2-17 x-31 y-7=0,

3 x+4 y-5=0

xy

5{x^2} + 4\sqrt 3 \;xy + 9{y^2} - 8 = 0

\alpha=\ldots \ldots

x \cos \alpha+y \sin \alpha=p

p \neq 0

\sqrt{3}

(a b /(a-b), 0)

(a-b)\left(x^2+y^2\right)-2 a b x-0

y^2+4 y+8 x-2=0

y