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Question

Through what angle should the axes be rotated so that equation ax2−2√3xy+7y2=10(x,y) may be changed to 3x2+5y2=5(x1,y1)

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Solution

Solution:-ax2−2√3xy+7y2=10(x,y)Let the axes rotated by an angle θ.Let the equation after rotation be-aX2−2√3XY+7Y2=10X=xcosθ+ysinθY=xsinθ−ycosθ∴aX2−2√3XY+7Y2=10⇒a(xcosθ+ysinθ)2−2√3(xcosθ+ysinθ)(xsinθ−ycosθ)+7(xsinθ−ycosθ)2=10⇒ax2cos2θ+ay2sin2θ+2axysinθcosθ−2√3x2cosθsinθ−2√3xysin2θ+2√3xycos2θ+2√3y2cosθsinθ+7x2sin2θ+7y2cos2θ−14xysinθcosθ=10⇒x2(acos2θ+7sin2θ−2√3sinθcosθ)+y2(asin2θ+7cos2θ+2√3cosθsinθ)+xy(2acosθsinθ−14sinθcosθ−2√3sin2θ+2√3cos2θ)=10⇒x2(acos2θ+7sin2θ−2√3sinθcosθ)+y2(asin2θ+7cos2θ+2√3cosθsinθ)+xy(asin2θ−7sin2θ+2√3(cos2θ−sin2θ))=10⇒x2(acos2θ+7sin2θ−2√3sinθcosθ)+y2(asin2θ+7cos2θ+2√3cosθsinθ)+xy(asin2θ−7sin2θ+2√3cos2θ)=10As in equation 3x2+5y2=5(x1,y1)Coefficient of xy is 0.∴asin2θ−7sin2θ+2√3cos2θ=0⇒sin2θ(a−7)=−2√3cos2θ⇒tan2θ=−2√3a−7⇒θ=tan−1(−2√3a−7)2

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