Given,
The equation
4x2+2√3xy+2y2−1=0
Let old coordinates (x,y)
new coordinates (X,Y)
∵x=Xcosθ−Ysinθ
y=Xsinθ+Ycosθ
Putting the value of (x) and (y) in equation (i)
4(xcosθ−ysinθ)2+2√3(xcosθ−ysinθ)(xsinθ+ycosθ)
+2(xsinθ+ycosθ)2−1=0
⇒4(x2cos2θ+y2sin2θ−2xycosθsinθ)+
2√3(x2cosθsinθ+xcos2θ−xysin2θ−y2sinθcosθ)
+2(x2sin2θ+y2cos2θ+2xysinθcosθ)−1=0
⇒x2(4cos2θ+2sin2θ+2√3cosθsinθ)+y2(4sin2θ−2√3sinθ
+2cos2θ)+xy(−8sincotθ+2√3cos2θ−2√3sin2θ
+4sinθcosθ)−1=0
equating this equation with 5x2+y2=1,
∴xy=0
⇒ -. −2√3(sin2θ−cos2θ)+−4sinθcosθ=0
⇒2√3cos2θ−2sin2θ=D
⇒2×2(√32cos2θ−12sin2θ)=0
⇒(sin60∘cos2θ−cos60∘sin2θ)=θ
⇒sin(60−2θ)=0
⇒60=2θ[∵sin0∘=0]
⇒θ=30∘
Option (B) is correct