Show that the line ` 3x+ sqrt3y =12` is a tangent to the ellipse ` 9x^(2) +y^(2) = 36` . Find the coordinates of the point of contact.

Answer

To show that the line \(3x + \sqrt{3}y = 12\) is a tangent to the ellipse \(9x^2 + y^2 = 36\) and to find the coordinates of the point of contact, we can follow these steps:

### Step 1: Rewrite the line equation in slope-intercept form
We start with the line equation:
\[ 3x + \sqrt{3}y = 12 \]
We can isolate \(y\):
\[ \sqrt{3}y = 12 - 3x \]
\[ y = \frac{12}{\sqrt{3}} - \frac{3}{\sqrt{3}}x \]
This simplifies to:
\[ y = 4\sqrt{3} - \sqrt{3}x \]
From this, we can identify:
- The slope \(m = -\sqrt{3}\)
- The y-intercept \(c = 4\sqrt{3}\)

### Step 2: Rewrite the ellipse equation in standard form
The equation of the ellipse is:
\[ 9x^2 + y^2 = 36 \]
Dividing the entire equation by 36 gives:
\[ \frac{x^2}{4} + \frac{y^2}{36} = 1 \]
From this, we identify:
- \(a^2 = 4 \Rightarrow a = 2\)
- \(b^2 = 36 \Rightarrow b = 6\)

### Step 3: Check the condition for tangency
The condition for a line \(y = mx + c\) to be tangent to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is given by:
\[ c^2 = a^2m^2 + b^2 \]
Substituting the values we have:
- \(c = 4\sqrt{3}\)
- \(m = -\sqrt{3}\)
- \(a^2 = 4\)
- \(b^2 = 36\)

Calculating \(c^2\):
\[ (4\sqrt{3})^2 = 48 \]

Calculating \(a^2m^2 + b^2\):
\[ 4 \cdot (-\sqrt{3})^2 + 36 = 4 \cdot 3 + 36 = 12 + 36 = 48 \]

Since both sides are equal:
\[ 48 = 48 \]
This confirms that the line is indeed tangent to the ellipse.

### Step 4: Find the coordinates of the point of contact
The coordinates of the point of contact can be found using the formula:
\[ \left(-\frac{a^2m}{c}, \frac{b^2}{c}\right) \]
Substituting the values:
\[ \left(-\frac{4 \cdot (-\sqrt{3})}{4\sqrt{3}}, \frac{36}{4\sqrt{3}}\right) \]
Calculating the x-coordinate:
\[ -\frac{4 \cdot (-\sqrt{3})}{4\sqrt{3}} = 1 \]

Calculating the y-coordinate:
\[ \frac{36}{4\sqrt{3}} = \frac{9}{\sqrt{3}} = 3\sqrt{3} \]

Thus, the coordinates of the point of contact are:
\[ (1, 3\sqrt{3}) \]

### Final Answer
The line \(3x + \sqrt{3}y = 12\) is a tangent to the ellipse \(9x^2 + y^2 = 36\), and the coordinates of the point of contact are \((1, 3\sqrt{3})\).