UPSC NDA I 2019 Mathematics Solved Paper

(B) Given, p = Xcosθ − Y sin θ q = X sin θ + Y cosθ p$^2$ + 4 pq + q$^2$ = A X$^2$ +BY$^2$ …(i) p$^2$ + 4 pq + q$^2$ = X$^2$ cos$^2$ θ + Y$^2$ sin$^2$θ − 2XY cosθ sin θ + x$^2$ sin$^2$θ + Y$^2$cos$^2$θ + 2 XY sin θ cosθ + 4[(Xcosθ− Ysinθ) (X sinθ + Y cosθ) ] = X$^2$ + Y$^2$ + 4 [ X$^2$ sin θ cosθ + XY cos$^2$θ − XY sin$^2$θ − Y$^2$sin θ cosθ ] = X$^2$ + Y$^2$ + 4$\left[ (X^2 - Y^2) \frac{\sin 2\theta}{2} + XY \cos 2\theta \right]$ = X$^2$ + Y$^2$ + 2( X$^2$ −Y$^2$)sin 2 θ + 4 XY cos 2θ …(ii) From (i) and (ii) we get cos2 θ = 0 ⇒ θ = $\frac{\pi}{4}$ ∴ X$^2$ + Y$^2$ + 2(X$^2$ − Y$^2$) = A X$^2$ + BY$^2$ 3 X$^2$ − Y$^2$ = A X$^2$ + BY$^2$ ⇒ A = 3 ; B = −1