$$ 4 ( ( \frac { \sqrt { 3 } } { 2 } x - \frac { 1 } { 2 } y ) ^ { 2 } ) + ( 2 \sqrt { 3 } ( \frac { \sqrt { 3 } } { 2 } x - \frac { 1 } { 2 } y ) ( \frac { 1 } { 2 } x + \frac { \sqrt { 3 } } { 2 } y ) ) + 2 ( ( \frac { 1 } { 2 } x + \frac { \sqrt { 3 } } { 2 } y ) ^ { 2 } ) + 10 \sqrt { 3 } ( \frac { \sqrt { 3 } } { 2 } x - \frac { 1 } { 2 } y ) + 10 ( \frac { 1 } { 2 } x + \frac { \sqrt { 3 } } { 2 } y ) = 5 $$

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UpStudy AI · Accepted Answer

Let's simplify the given equation step by step. The original equation is:

$$
4 \left( \left( \frac{\sqrt{3}}{2} x - \frac{1}{2} y \right)^2 \right) + 2\sqrt{3} \left( \left( \frac{\sqrt{3}}{2} x - \frac{1}{2} y \right) \left( \frac{1}{2} x + \frac{\sqrt{3}}{2} y \right) \right) + 2 \left( \left( \frac{1}{2} x + \frac{\sqrt{3}}{2} y \right)^2 \right) + 10\sqrt{3} \left( \frac{\sqrt{3}}{2} x - \frac{1}{2} y \right) + 10 \left( \frac{1}{2} x + \frac{\sqrt{3}}{2} y \right) = 5
$$

### Step 1: Introduce Substitutions
Let’s define:
$$
A = \frac{\sqrt{3}}{2}x - \frac{1}{2}y
$$
$$
B = \frac{1}{2}x + \frac{\sqrt{3}}{2}y
$$

This allows us to rewrite the equation as:
$$
4A^2 + 2\sqrt{3}AB + 2B^2 + 10\sqrt{3}A + 10B = 5
$$

### Step 2: Expand and Simplify

1. **Expand $$4A^2$$:**
   $$
   4A^2 = 4\left(\frac{\sqrt{3}}{2}x - \frac{1}{2}y\right)^2 = 3x^2 - 2\sqrt{3}xy + y^2
   $$

2. **Expand $$2\sqrt{3}AB$$:**
   $$
   AB = \left(\frac{\sqrt{3}}{2}x - \frac{1}{2}y\right)\left(\frac{1}{2}x + \frac{\sqrt{3}}{2}y\right) = \frac{\sqrt{3}}{4}x^2 + \frac{1}{2}xy - \frac{\sqrt{3}}{4}y^2
   $$
   $$
   2\sqrt{3}AB = \frac{3}{2}x^2 + \sqrt{3}xy - \frac{3}{2}y^2
   $$

3. **Expand $$2B^2$$:**
   $$
   2B^2 = 2\left(\frac{1}{2}x + \frac{\sqrt{3}}{2}y\right)^2 = \frac{1}{2}x^2 + \sqrt{3}xy + \frac{3}{2}y^2
   $$

4. **Combine the Quadratic Terms:**
   $$
   4A^2 + 2\sqrt{3}AB + 2B^2 = \left(3x^2 - 2\sqrt{3}xy + y^2\right) + \left(\frac{3}{2}x^2 + \sqrt{3}xy - \frac{3}{2}y^2\right) + \left(\frac{1}{2}x^2 + \sqrt{3}xy + \frac{3}{2}y^2\right)
   $$
   $$
   = 5x^2 + y^2
   $$

5. **Expand the Linear Terms:**
   $$
   10\sqrt{3}A = 10\sqrt{3}\left(\frac{\sqrt{3}}{2}x - \frac{1}{2}y\right) = 15x - 5\sqrt{3}y
   $$
   $$
   10B = 10\left(\frac{1}{2}x + \frac{\sqrt{3}}{2}y\right) = 5x + 5\sqrt{3}y
   $$
   $$
   10\sqrt{3}A + 10B = 20x
   $$

6. **Combine All Terms:**
   $$
   5x^2 + y^2 + 20x = 5
   $$

### Step 3: Complete the Square

1. **Focus on the $$x$$-terms:**
   $$
   5x^2 + 20x = 5(x^2 + 4x)
   $$
   $$
   = 5\left[(x)^2 + 4x + 4 - 4\right] = 5\left[(x + 2)^2 - 4\right] = 5(x + 2)^2 - 20
   $$

2. **Substitute Back:**
   $$
   5(x + 2)^2 - 20 + y^2 = 5
   $$
   $$
   5(x + 2)^2 + y^2 = 25
   $$

3. **Divide Both Sides by 25 to Get the Standard Form:**
   $$
   \frac{(x + 2)^2}{5} + \frac{y^2}{25} = 1
   $$

### Conclusion

The given equation simplifies to the standard form of an ellipse:

$$
\boxed{ \frac{(x + 2)^2}{5} + \frac{y^2}{25} = 1 }
$$

This represents an ellipse centered at $$(-2, 0)$$ with a semi-major axis of length $$5$$ along the $$y$$-axis and a semi-minor axis of length $$\sqrt{5}$$ along the $$x$$-axis.
The equation simplifies to the ellipse: $$ \frac{(x + 2)^2}{5} + \frac{y^2}{25} = 1 $$

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