Answer to Question #114930 in Analytic Geometry for ANJU JAYACHANDRAN

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 45^â—¦
, followed by the shifting of the origin at
(2,−1)?

Answer 1

Given, the equation of conic is $x^2+y^2+4x-2y+3=0 \: (\clubs)$ .

i). Now, suppose the conics axes are rotated by $45^{\circ}$ and let us consider new axes are $x' \&y'$ ,thus

$\begin{bmatrix} x' \\ y' \end{bmatrix}=\begin{bmatrix} \cos(45^{\circ}) & \sin(45^{\circ}) \\ -\sin(45^{\circ})&\cos(45^{\circ}) \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix} \implies \begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} \cos(45^{\circ}) & -\sin(45^{\circ}) \\ \sin(45^{\circ})&\cos(45^{\circ}) \end{bmatrix}\begin{bmatrix} x'\\ y' \end{bmatrix}$ ,

moreover

$$x=\frac{x'-y'}{\sqrt{2}} \: (\star)\\ y=\frac{x'+y'}{\sqrt{2}} \: (\star \star)$$

On solving the equation $(\clubs)$ ,we get,

new equation of conic is $(\frac{x'-y'}{\sqrt{2}})^2+(\frac{x'+y'}{\sqrt{2}})^2 +4(\frac{x'-y'}{\sqrt{2}})-2(\frac{x'+y'}{\sqrt{2}})+3=0$ , On simplification equation of conic becomes,

$$\sqrt{2}(x'^2+y'^2)+(2x' -6y')+3\sqrt{2}=0 \: (\dag)$$

Now , if origin is shifted by

$$(2,-1) \implies x'=x''-2 \: \& \: y'=y''+1 \: (\diamond)$$

,hence from the $(\dag) \: \& (\diamond)$ we get $\sqrt{2}\{(x'' -2)^2+(y''+1)^2)\}+2(x''-2) -6(y'+1')+3\sqrt{2}=0$ ,thus on simplification we get the final equation of conic is

$$\sqrt{2}(x^2+y^2)+(2-4\sqrt{2})x+(2\sqrt{2} -6)y+8\sqrt{2}=0$$

As, $x'',x' \dots y' \&x,y$ are dummy variable.

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 $x'' \leftrightarrow x' \: \& y'' \leftrightarrow y'$ ,we get, $\big(\square )$

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

Now rotate the axes as part (i) equation $(\star), (\star \star)$ and considering $\big(\square)$ , thus

$$(\frac{x''-y''}{\sqrt{2}})^2+(\frac{x''+y''}{\sqrt{2}})^2 -2=0\\ \implies x''^2+y''^2-2=0$$

Therefore, the final equation of the conic is

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

Hence, we are done.