Answer to Question #114931 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^◦
, followed by the shifting of the origin at
(2,−1)?
Are the equations in i) and ii) above the same? Why?

Answer 1

Solution.

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

i)the origin is shifted at (2,−1)

$x'=x+2; y'=y-1;$

$(x+2)^2+(y-1)^2+4(x+2)-2(y-1)+3=0;$

$x^2+4x+4+y^2-2y+1+4x+8-2y+2+3=0;$

$x^2+y^2+8x-4y+18=0;$

the axes are rotated through 45^o;

$x'=xcos\alpha-ysin\alpha; y'=xsin\alpha+ycos\alpha;$

$x'=x\dfrac{\sqrt{2}}{2}-y\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{2}(x-y);$

$y'=x\dfrac{\sqrt{2}}{2}+y\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{2}(x+y);$

$\dfrac{1}{2}(x-y)^2+\dfrac{1}{2}(x+y)^2+4\sqrt{2}(x-y)-2\sqrt{2}(x+y)+18=0;$

$\dfrac{1}{2}x^2-xy+\dfrac{1}{2}y^2+\dfrac{1}{2}x^2+xy+\dfrac{1}{2}y^2+4\sqrt{2}x-4\sqrt{2}y- 2\sqrt{2}x-2\sqrt{2}y+18=0;$

$x^2+y^2+2\sqrt{2}x-6\sqrt{2}y+18=0;$

ii)the axes are rotated through 45^o;

$x'=xcos\alpha-ysin\alpha; y'=xsin\alpha+ycos\alpha;$

$x'=x\dfrac{\sqrt{2}}{2}-y\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{2}(x-y);$

$y'=x\dfrac{\sqrt{2}}{2}+y\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{2}(x+y);$

$\dfrac{1}{2}(x-y)^2+\dfrac{1}{2}(x+y)^2+2\sqrt{2}(x-y)-\sqrt{2}(x+y)+3=0;$

$\dfrac{1}{2}x^2-xy+\dfrac{1}{2}y^2+\dfrac{1}{2}x^2+xy+\dfrac{1}{2}y^2+2\sqrt{2}x-2\sqrt{2}y- \sqrt{2}x-\sqrt{2}y+3=0;$

$x^2+y^2+\sqrt{2}x-3\sqrt{2}y+3=0;$

the origin is shifted at (2,−1)

$x'=x+2; y'=y-1;$

$(x+2)^2+(y-1)^2+\sqrt{2}(x+2)-3\sqrt{2}(y-1)+3=0;$

$x^2+4x+4+y^2-2y+1+\sqrt{2}x+2\sqrt{2}-3\sqrt{2}y+3\sqrt{2}+3=0;$

$x^2+y^2+x(4+\sqrt{2})-y(2+3\sqrt{2})+8+5\sqrt{2}=0;$

Answer: i)$x^2+y^2+2\sqrt{2}x-6\sqrt{2}y+18=0;$

ii)$x^2+y^2+x(4+\sqrt{2})-y(2+3\sqrt{2})+8+5\sqrt{2}=0;$

the equations i) and ii) are different.