For which of the following curves, the line `x + sqrt3 y = 2sqrt3` is the tangent at the point `((3sqrt3)/(2), 1/2)` ?

Options

A. `x^2 + y^2 = 7`
B. `y^2=(1)/(6sqrt3)x`
C. `2x^2 - 18y^2 = 9`
D. `x^2+9y^2 = 9`

Answer

To determine which of the given curves has the line \( x + \sqrt{3} y = 2\sqrt{3} \) as a tangent at the point \( \left( \frac{3\sqrt{3}}{2}, \frac{1}{2} \right) \), we will analyze each option step by step.

### Step 1: Identify the Tangent Line
The equation of the tangent line is given as:
\[ x + \sqrt{3} y = 2\sqrt{3} \]
We can rewrite it in slope-intercept form:
\[ \sqrt{3} y = -x + 2\sqrt{3} \implies y = -\frac{1}{\sqrt{3}} x + \frac{2}{\sqrt{3}} \]
The slope of the tangent line is \( -\frac{1}{\sqrt{3}} \).

### Step 2: Check Each Curve Option
We will check each option to see if the tangent line is valid at the given point.

#### Option A: \( x^2 + 9y^2 = 9 \)
1. Differentiate the equation implicitly:
\[ 2x + 18y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{2x}{18y} = -\frac{x}{9y} \]
2. Substitute the point \( \left( \frac{3\sqrt{3}}{2}, \frac{1}{2} \right) \):
\[ \frac{dy}{dx} = -\frac{\frac{3\sqrt{3}}{2}}{9 \cdot \frac{1}{2}} = -\frac{3\sqrt{3}}{9} = -\frac{\sqrt{3}}{3} \]
This slope matches \( -\frac{1}{\sqrt{3}} \) (as \( -\frac{\sqrt{3}}{3} = -\frac{1}{\sqrt{3}} \)), so this option is valid.

#### Option B: \( 2x^2 + 18y^2 = 9 \)
1. Differentiate:
\[ 4x + 36y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{4x}{36y} = -\frac{x}{9y} \]
2. Substitute the point:
\[ \frac{dy}{dx} = -\frac{\frac{3\sqrt{3}}{2}}{9 \cdot \frac{1}{2}} = -\frac{3\sqrt{3}}{9} = -\frac{\sqrt{3}}{3} \]
This slope also matches \( -\frac{1}{\sqrt{3}} \), so this option is valid.

#### Option C: \( y^2 = 6\sqrt{3}x \)
1. Differentiate:
\[ 2y \frac{dy}{dx} = 6\sqrt{3} \implies \frac{dy}{dx} = \frac{6\sqrt{3}}{2y} = \frac{3\sqrt{3}}{y} \]
2. Substitute the point:
\[ \frac{dy}{dx} = \frac{3\sqrt{3}}{\frac{1}{2}} = 6\sqrt{3} \]
This slope does not match \( -\frac{1}{\sqrt{3}} \), so this option is invalid.

#### Option D: \( x^2 + y^2 = 7 \)
1. Differentiate:
\[ 2x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y} \]
2. Substitute the point:
\[ \frac{dy}{dx} = -\frac{\frac{3\sqrt{3}}{2}}{\frac{1}{2}} = -3\sqrt{3} \]
This slope does not match \( -\frac{1}{\sqrt{3}} \), so this option is invalid.

### Conclusion
The curves for which the line \( x + \sqrt{3} y = 2\sqrt{3} \) is the tangent at the point \( \left( \frac{3\sqrt{3}}{2}, \frac{1}{2} \right) \) are:
- Option A: \( x^2 + 9y^2 = 9 \)
- Option B: \( 2x^2 + 18y^2 = 9 \)