Answer to Question #104250 in Analytic Geometry for Deepak Rana

Question

What is the new equation of the conic x
2 +y
2 +4x−2y+3 = 0, when
i) the origin is shifted at (2,−1), followed by a rotation of axes through 45◦
?
ii) the axes are rotated through 45â—¦
, followed by the shifting of the origin at
(2,−1)?
Are the equations in i) and ii) above the same? Why?

Answer 1

Let

$x^2+y^2+4x-2y+3=0$

i) the origin is shifted at (2,−1)

$x=x'+a=x'+2\\ y=y'+b=y'-1$

substitute into the equation

$(x'+2)^2+(y'-1)^2+4(x'+2)-2(y'-1)+3=0\\ x'^2+4x'+4+y'^2-2y'+1+4x'+8-2y'+2+3=0\\ x'^2+8x'+y'^2-4y'+18=0$

 the axes are rotated through $45^0$

$x'=Xcos\alpha-Ysin\alpha=Xcos45^0-Ysin45^0=\\ =\frac{\sqrt2}{2}X-\frac{\sqrt2}{2}Y=\frac{\sqrt2}{2}(X-Y)\\ y'=Xsin\alpha+Ycos\alpha=Xsin45^0+Ycos45^0=\\ =\frac{\sqrt2}{2}X+\frac{\sqrt2}{2}Y=\frac{\sqrt2}{2}(X+Y)\\$

substitute into the equation

$\frac{2}{4}(X-Y)^2+8\cdot\frac{\sqrt2}{2}(X-Y)+\\ +\frac{2}{4}(X+Y)^2-4\cdot\frac{\sqrt2}{2}(X+Y)+18=0\\ \frac{1}{2}(X^2-2XY+Y^2)+4\sqrt2(X-Y)+\\ +\frac{1}{2}(X^2+2XY+Y^2)-2\sqrt2(X+Y)+18=0\\ X^2+Y^2+2\sqrt2X-6\sqrt2Y+18=0$

ii)  the axes are rotated through $45^0$

$x=x'cos\alpha-y'sin\alpha=x'cos45^0-y'sin45^0=\frac{\sqrt2}{2}(x'-y')\\ y=x'sin\alpha+y'cos\alpha=x'sin45^0+y'cos45^0=\frac{\sqrt2}{2}(x'+y')\\$

substitute into the equation

$\frac{1}{2}(x'-y')^2+\frac{1}{2}(x'+y')^2+\\ +4\cdot\frac{\sqrt2}{2}(x'-y)-2\cdot\frac{\sqrt2}{2}(x'+y')+3=0\\ \frac{1}{2}(x'^2-2x'y'+y'^2)+\frac{1}{2}(x'^2+2x'y'+y'^2)+\\ +2\sqrt2(x'-y')-\sqrt2(x'+y')+3=0\\ x'^2+y'^2+\sqrt2x'-3\sqrt2y'+3=0$

the origin is shifted at (2,−1)

$x'=X+a=X+2\\ y'=Y+b=Y-1$

substitute into the equation

$(X+2)^2+(Y-1)^2+\sqrt2(X+2)-3\sqrt2(Y-1)+3=0\\ X^2+4X+4+Y^2-2Y+1+\sqrt2X+\\ +2\sqrt2-3\sqrt2Y+3\sqrt2+3=0\\ X^2+Y^2+X(4+\sqrt2)-Y(2+3\sqrt2)+8+5\sqrt2=0$

the equations  i) and ii) are different