Which one of the following points does not lie on the normal to the hyperbol `(x^(2))/(16)+(y^(2))/(9)` =1 drawn at the point `(8,3sqrt(3))`

Options

A. `10,(1)/sqrt(3)`
B. `13,(1)/sqrt(3)`
C. `12,(1)/sqrt(3)`
D. `11,sqrt(3)`

Answer

To solve the problem, we need to determine which point does not lie on the normal to the hyperbola \(\frac{x^2}{16} + \frac{y^2}{9} = 1\) drawn at the point \((8, 3\sqrt{3})\).

### Step-by-Step Solution:

1. **Find the slope of the tangent line at the given point**:
The equation of the hyperbola is \(\frac{x^2}{16} + \frac{y^2}{9} = 1\). To find the slope of the tangent line at the point \((8, 3\sqrt{3})\), we first differentiate the equation implicitly.

\[ \frac{d}{dx}\left(\frac{x^2}{16} + \frac{y^2}{9}\right) = 0 \]

This gives us:

\[ \frac{2x}{16} + \frac{2y}{9} \cdot \frac{dy}{dx} = 0 \]

Rearranging, we find:

\[ \frac{dy}{dx} = -\frac{9x}{16y} \]

Substituting \(x = 8\) and \(y = 3\sqrt{3}\):

\[ \frac{dy}{dx} = -\frac{9 \cdot 8}{16 \cdot 3\sqrt{3}} = -\frac{72}{48\sqrt{3}} = -\frac{3}{2\sqrt{3}} = -\frac{2}{\sqrt{3}} \]

2. **Find the slope of the normal line**:
The slope of the normal line is the negative reciprocal of the slope of the tangent line:

\[ m_{\text{normal}} = -\frac{1}{-\frac{2}{\sqrt{3}}} = \frac{\sqrt{3}}{2} \]

3. **Write the equation of the normal line**:
Using the point-slope form of the line equation:

\[ y - y_1 = m(x - x_1) \]

where \((x_1, y_1) = (8, 3\sqrt{3})\):

\[ y - 3\sqrt{3} = \frac{\sqrt{3}}{2}(x - 8) \]

Rearranging gives:

\[ y = \frac{\sqrt{3}}{2}x - 4\sqrt{3} + 3\sqrt{3} \]

\[ y = \frac{\sqrt{3}}{2}x - \sqrt{3} \]

4. **Convert to standard form**:
To make it easier to check points, we can rearrange this equation:

\[ \sqrt{3}y = \frac{\sqrt{3}}{2}x - \sqrt{3} \]

Multiplying through by 2 to eliminate the fraction:

\[ 2\sqrt{3}y = \sqrt{3}x - 2\sqrt{3} \]

Rearranging gives:

\[ \sqrt{3}x - 2\sqrt{3}y - 2\sqrt{3} = 0 \]

5. **Check the given points**:
Now we need to check which of the given points does not satisfy the equation of the normal line. Let's denote the points as \(A(10, \frac{1}{\sqrt{3}})\), \(B(13, \frac{1}{\sqrt{3}})\), and \(C(12, \frac{1}{\sqrt{3}})\).

- For point \(A(10, \frac{1}{\sqrt{3}})\):

\[ \sqrt{3}(10) - 2\sqrt{3}\left(\frac{1}{\sqrt{3}}\right) - 2\sqrt{3} = 10\sqrt{3} - 2 - 2\sqrt{3} = 8\sqrt{3} - 2 \neq 0 \]

- For point \(B(13, \frac{1}{\sqrt{3}})\):

\[ \sqrt{3}(13) - 2\sqrt{3}\left(\frac{1}{\sqrt{3}}\right) - 2\sqrt{3} = 13\sqrt{3} - 2 - 2\sqrt{3} = 11\sqrt{3} - 2 \neq 0 \]

- For point \(C(12, \frac{1}{\sqrt{3}})\):

\[ \sqrt{3}(12) - 2\sqrt{3}\left(\frac{1}{\sqrt{3}}\right) - 2\sqrt{3} = 12\sqrt{3} - 2 - 2\sqrt{3} = 10\sqrt{3} - 2 \neq 0 \]

After checking, we find that point \(A(10, \frac{1}{\sqrt{3}})\) does not satisfy the equation of the normal line.

### Conclusion:
The point that does not lie on the normal to the hyperbola at the given point is \(A(10, \frac{1}{\sqrt{3}})\).