Question

Q3: Identify the graph of the equation and write and equation of the translated or rotated graph in general form. (Picture Provided Below)

Answer 1

Answer:

b. circle; $2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

Step-by-step explanation:

The given conic has equation;

$x^2-5x+y^2=3$

We complete the square to obtain;

$(x-\frac{5}{2})^2+(y-0)^2=\frac{37}{4}$

This is a circle with center;

$(\frac{5}{2},0)$

This implies that;

$x=\frac{5}{2},y=0$

When the circle is rotated through an angle of $\theta=\frac{\pi}{3}$,

The new center is obtained using;

$x'=x\cos(\theta)+y\sin(\theta)$ and $y'=-x\sin(\theta)+y\cos(\theta)$

We plug in the given angle with x and y values to get;

$x'=(\frac{5}{2})\cos(\frac{\pi}{3})+(0)\sin(\frac{\pi}{3})$ and $y'=--(\frac{5}{2})\sin(\frac{\pi}{3})+(0)\cos(\frac{\pi}{3})$

This gives us;

$x'=\frac{5}{4} ,y'=\frac{5\sqrt{3} }{4}$

The equation of the rotated circle is;

$(x'-\frac{5}{4})^2+(y'-\frac{5\sqrt{3} }{4})^2=\frac{37}{4}$

Expand;

$(x')^2+(y')^2-\frac{5\sqrt{3} }{2}y'-\frac{5}{2}x'+\frac{25}{4} =\frac{37}{4}$

Multiply through by 4; to get

$4(x')^2+4(y')^2-10\sqrt{3}y'-10x'+25 =37$

Write in general form;

$4(x')^2+4(y')^2-10x'-10\sqrt{3}y'-12 =0$

Divide through by 2.

$2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$