JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

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Application +4 | -1

0 \leq \theta \leq \frac{\pi}{2}

and

x=X \cos \theta+Y \sin \theta

y=X \sin \theta-Y \cos \theta

such that

x^2+2 x y+y^2=a X^2+b Y^2

, where

a

and

b

are constants, then

View answer and explanation

Correct option: C

a=3, b=-1

Option C (

a=3, b=-1

) is correct because by substituting the given expressions for x and y into the equation and comparing coefficients, we find that

a=3

and

b=-1

when

\theta = \frac{{\pi}}{{4}}

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