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arsen [322]
3 years ago
7

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

ded Below)

Mathematics
1 answer:
natta225 [31]3 years ago
8 0

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

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