Let angle be θ then replacing (x,y) by (x‌cos‌θ−ysin‌θ,xsin‌θ+y‌cos‌θ) then 9x2−2√3xy+7y2=10 becomes ‌9(x‌cos‌θ−ysin‌θ)2−2√3(x‌cos‌θ−ysin‌θ) ‌(xsin‌θ+y‌cos‌θ)+7(xsin‌θ+y‌cos‌θ)2=10 ‌⇒x2(9cos2θ−sin‌θ‌cos‌θ+7sin‌2θ)+2xy(−9sin‌θ ‌cos‌θ−√3‌cos‌2‌θ+7sin‌θ‌cos‌θ) ‌+y2(9cos2θ+sin‌θ‌cos‌θ+7cos2θ)=10 On comparing with 3x2+5y2=5 (coefficient of xy=0 ) We get −9sin‌θ‌cos‌θ−√3‌cos‌2‌θ+7sin‌θ‌cos‌θ=0 or sin‌2θ=−√3‌cos‌2‌θ or tan‌2‌θ=−√3=tan(180∘−60∘) or 2θ=120∘ ∴‌‌θ=60∘
