Answer:
b. circle; [tex]2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0[/tex]
Step-by-step explanation:
The given conic has equation;
[tex]x^2-5x+y^2=3[/tex]
We complete the square to obtain;
[tex](x-\frac{5}{2})^2+(y-0)^2=\frac{37}{4}[/tex]
This is a circle with center;
[tex](\frac{5}{2},0)[/tex]
This implies that;
[tex]x=\frac{5}{2},y=0[/tex]
When the circle is rotated through an angle of [tex]\theta=\frac{\pi}{3}[/tex],
The new center is obtained using;
[tex]x'=x\cos(\theta)+y\sin(\theta)[/tex] and [tex]y'=-x\sin(\theta)+y\cos(\theta)[/tex]
We plug in the given angle with x and y values to get;
[tex]x'=(\frac{5}{2})\cos(\frac{\pi}{3})+(0)\sin(\frac{\pi}{3})[/tex] and [tex]y'=--(\frac{5}{2})\sin(\frac{\pi}{3})+(0)\cos(\frac{\pi}{3})[/tex]
This gives us;
[tex]x'=\frac{5}{4} ,y'=\frac{5\sqrt{3} }{4}[/tex]
The equation of the rotated circle is;
[tex](x'-\frac{5}{4})^2+(y'-\frac{5\sqrt{3} }{4})^2=\frac{37}{4}[/tex]
Expand;
[tex](x')^2+(y')^2-\frac{5\sqrt{3} }{2}y'-\frac{5}{2}x'+\frac{25}{4} =\frac{37}{4}[/tex]
Multiply through by 4; to get
[tex]4(x')^2+4(y')^2-10\sqrt{3}y'-10x'+25 =37[/tex]
Write in general form;
[tex]4(x')^2+4(y')^2-10x'-10\sqrt{3}y'-12 =0[/tex]
Divide through by 2.
[tex]2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0[/tex]
