Given, the equation of conic is x 2 + y 2 + 4 x − 2 y + 3 = 0 ( ♣ ) x^2+y^2+4x-2y+3=0 \: (\clubs) x 2 + y 2 + 4 x − 2 y + 3 = 0 ( ♣ )
i). Now, suppose the conics axes are rotated by 4 5 ∘ 45^{\circ} 4 5 ∘ x ′ & y ′ x' \&y' x ′ & y ′
[ x ′ y ′ ] = [ cos ( 4 5 ∘ ) sin ( 4 5 ∘ ) − sin ( 4 5 ∘ ) cos ( 4 5 ∘ ) ] [ x y ] ⟹ [ x y ] = [ cos ( 4 5 ∘ ) − sin ( 4 5 ∘ ) sin ( 4 5 ∘ ) cos ( 4 5 ∘ ) ] [ x ′ y ′ ] \begin{bmatrix}
x' \\
y'
\end{bmatrix}=\begin{bmatrix}
\cos(45^{\circ}) & \sin(45^{\circ}) \\
-\sin(45^{\circ})&\cos(45^{\circ})
\end{bmatrix}\begin{bmatrix}
x\\
y
\end{bmatrix} \implies \begin{bmatrix}
x \\
y
\end{bmatrix}=\begin{bmatrix}
\cos(45^{\circ}) & -\sin(45^{\circ}) \\
\sin(45^{\circ})&\cos(45^{\circ})
\end{bmatrix}\begin{bmatrix}
x'\\
y'
\end{bmatrix} [ x ′ y ′ ] = [ cos ( 4 5 ∘ ) − sin ( 4 5 ∘ ) sin ( 4 5 ∘ ) cos ( 4 5 ∘ ) ] [ x y ] ⟹ [ x y ] = [ cos ( 4 5 ∘ ) sin ( 4 5 ∘ ) − sin ( 4 5 ∘ ) cos ( 4 5 ∘ ) ] [ x ′ y ′ ]
moreover
x = x ′ − y ′ 2 ( ⋆ ) y = x ′ + y ′ 2 ( ⋆ ⋆ ) x=\frac{x'-y'}{\sqrt{2}} \: (\star)\\ y=\frac{x'+y'}{\sqrt{2}} \: (\star \star) x = 2 x ′ − y ′ ( ⋆ ) y = 2 x ′ + y ′ ( ⋆ ⋆ ) On solving the equation ( ♣ ) (\clubs) ( ♣ )
new equation of conic is ( x ′ − y ′ 2 ) 2 + ( x ′ + y ′ 2 ) 2 + 4 ( x ′ − y ′ 2 ) − 2 ( x ′ + y ′ 2 ) + 3 = 0 (\frac{x'-y'}{\sqrt{2}})^2+(\frac{x'+y'}{\sqrt{2}})^2 +4(\frac{x'-y'}{\sqrt{2}})-2(\frac{x'+y'}{\sqrt{2}})+3=0 ( 2 x ′ − y ′ ) 2 + ( 2 x ′ + y ′ ) 2 + 4 ( 2 x ′ − y ′ ) − 2 ( 2 x ′ + y ′ ) + 3 = 0
2 ( x ′ 2 + y ′ 2 ) + ( 2 x ′ − 6 y ′ ) + 3 2 = 0 ( † ) \sqrt{2}(x'^2+y'^2)+(2x' -6y')+3\sqrt{2}=0 \: (\dag) 2 ( x ′2 + y ′2 ) + ( 2 x ′ − 6 y ′ ) + 3 2 = 0 ( † ) Now , if origin is shifted by
( 2 , − 1 ) ⟹ x ′ = x ′ ′ − 2 & y ′ = y ′ ′ + 1 ( ⋄ ) (2,-1) \implies x'=x''-2 \: \& \: y'=y''+1 \: (\diamond) ( 2 , − 1 ) ⟹ x ′ = x ′′ − 2 & y ′ = y ′′ + 1 ( ⋄ ) ,hence from the ( † ) & ( ⋄ ) (\dag) \: \& (\diamond) ( † ) & ( ⋄ ) 2 { ( x ′ ′ − 2 ) 2 + ( y ′ ′ + 1 ) 2 ) } + 2 ( x ′ ′ − 2 ) − 6 ( y ′ + 1 ′ ) + 3 2 = 0 \sqrt{2}\{(x'' -2)^2+(y''+1)^2)\}+2(x''-2) -6(y'+1')+3\sqrt{2}=0 2 {( x ′′ − 2 ) 2 + ( y ′′ + 1 ) 2 )} + 2 ( x ′′ − 2 ) − 6 ( y ′ + 1 ′ ) + 3 2 = 0
2 ( x 2 + y 2 ) + ( 2 − 4 2 ) x + ( 2 2 − 6 ) y + 8 2 = 0 \sqrt{2}(x^2+y^2)+(2-4\sqrt{2})x+(2\sqrt{2} -6)y+8\sqrt{2}=0 2 ( x 2 + y 2 ) + ( 2 − 4 2 ) x + ( 2 2 − 6 ) y + 8 2 = 0
As, x ′ ′ , x ′ … y ′ & x , y x'',x' \dots y' \&x,y x ′′ , x ′ … y ′ & x , y
ii). Now, if we shift the origin first and then rotate the axis then exactly similar to above part from( ⋄ ) & ( ♣ ) (\diamond) \: \& \: (\clubs) ( ⋄ ) & ( ♣ )
if we replace the dummy variable x ′ ′ ↔ x ′ & y ′ ′ ↔ y ′ x'' \leftrightarrow x' \: \& y'' \leftrightarrow y' x ′′ ↔ x ′ & y ′′ ↔ y ′ ( □ ) \big(\square ) ( □ )
( x ′ − 2 ) 2 + ( y ′ + 1 ) 2 + 4 ( x ′ − 2 ) − 2 ( y ′ + 1 ) + 3 = 0 ⟹ x ′ 2 + y ′ 2 − 2 = 0 (x'-2)^2+(y'+1)^2 +4(x'-2)-2(y'+1) +3=0\\
\implies x'^2+y'^2-2=0 ( x ′ − 2 ) 2 + ( y ′ + 1 ) 2 + 4 ( x ′ − 2 ) − 2 ( y ′ + 1 ) + 3 = 0 ⟹ x ′2 + y ′2 − 2 = 0
Now rotate the axes as part (i) equation ( ⋆ ) , ( ⋆ ⋆ ) (\star), (\star \star) ( ⋆ ) , ( ⋆ ⋆ ) ( □ ) \big(\square) ( □ )
( x ′ ′ − y ′ ′ 2 ) 2 + ( x ′ ′ + y ′ ′ 2 ) 2 − 2 = 0 ⟹ x ′ ′ 2 + y ′ ′ 2 − 2 = 0 (\frac{x''-y''}{\sqrt{2}})^2+(\frac{x''+y''}{\sqrt{2}})^2 -2=0\\
\implies x''^2+y''^2-2=0 ( 2 x ′′ − y ′′ ) 2 + ( 2 x ′′ + y ′′ ) 2 − 2 = 0 ⟹ x ′′2 + y ′′2 − 2 = 0 Therefore, the final equation of the conic is
x 2 + y 2 − 2 = 0 x^2+y^2-2=0 x 2 + y 2 − 2 = 0 Hence, we are done.
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