Answer :
Let's find both the tangent line and the normal line to the graph of the function [tex]\( x^2 - \sqrt{3} x y + 4 y^2 = 235 \)[/tex] at the point [tex]\( (\sqrt{3}, 8) \)[/tex].
### Step-by-Step Solution:
1. Find the partial derivatives of the given equation with respect to [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
Given: [tex]\( f(x, y) = x^2 - \sqrt{3} x y + 4 y^2 - 235 \)[/tex]
The partial derivative with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{\partial f}{\partial x} = 2x - \sqrt{3} y \][/tex]
The partial derivative with respect to [tex]\(y\)[/tex]:
[tex]\[ \frac{\partial f}{\partial y} = -\sqrt{3} x + 8y \][/tex]
2. Evaluate the partial derivatives at the given point [tex]\( (\sqrt{3}, 8) \)[/tex]:
[tex]\[ \left. \frac{\partial f}{\partial x} \right|_{(\sqrt{3}, 8)} = 2(\sqrt{3}) - \sqrt{3} \cdot 8 = 2\sqrt{3} - 8\sqrt{3} = -6\sqrt{3} \][/tex]
[tex]\[ \left. \frac{\partial f}{\partial y} \right|_{(\sqrt{3}, 8)} = -\sqrt{3} \cdot \sqrt{3} + 8 \cdot 8 = -3 + 64 = 61 \][/tex]
3. Calculate the slope of the tangent line at the point [tex]\( (\sqrt{3}, 8) \)[/tex]:
The slope [tex]\( m_{\text{tangent}} \)[/tex] of the tangent line is given by:
[tex]\[ m_{\text{tangent}} = -\frac{\left. \frac{\partial f}{\partial x} \right|_{(\sqrt{3}, 8)}}{\left. \frac{\partial f}{\partial y} \right|_{(\sqrt{3}, 8)}} = -\frac{-6\sqrt{3}}{61} = \frac{6\sqrt{3}}{61} \][/tex]
4. Form the equation of the tangent line:
The tangent line passes through the point [tex]\( (\sqrt{3}, 8) \)[/tex] and has the slope [tex]\( \frac{6\sqrt{3}}{61} \)[/tex]. Using the point-slope form of a line:
[tex]\[ y - 8 = \frac{6\sqrt{3}}{61} (x - \sqrt{3}) \][/tex]
So, the equation of the tangent line is:
[tex]\[ y - 8 = \frac{6\sqrt{3}}{61} (x - \sqrt{3}) \][/tex]
5. Calculate the slope of the normal line:
The slope [tex]\( m_{\text{normal}} \)[/tex] of the normal line is the negative reciprocal of the slope of the tangent line:
[tex]\[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{\frac{6\sqrt{3}}{61}} = \frac{61}{6\sqrt{3}} \][/tex]
6. Form the equation of the normal line:
The normal line passes through the point [tex]\( (\sqrt{3}, 8) \)[/tex] and has the slope [tex]\( \frac{61}{6\sqrt{3}} \)[/tex]. Using the point-slope form of a line:
[tex]\[ y - 8 = \frac{61}{6\sqrt{3}} (x - \sqrt{3}) \][/tex]
So, the equation of the normal line is:
[tex]\[ y - 8 = \frac{61}{6\sqrt{3}} (x - \sqrt{3}) \][/tex]
### Final Equations:
- The equation of the tangent line is:
[tex]\[ y - 8 = \frac{6\sqrt{3}}{61} (x - \sqrt{3}) \][/tex]
- The equation of the normal line is:
[tex]\[ y - 8 = \frac{61}{6\sqrt{3}} (x - \sqrt{3}) \][/tex]
So, the normal line is:
[tex]\[ y-8 = \frac{61}{6\sqrt{3}}(x-\sqrt{3}) \][/tex]
### Step-by-Step Solution:
1. Find the partial derivatives of the given equation with respect to [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
Given: [tex]\( f(x, y) = x^2 - \sqrt{3} x y + 4 y^2 - 235 \)[/tex]
The partial derivative with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{\partial f}{\partial x} = 2x - \sqrt{3} y \][/tex]
The partial derivative with respect to [tex]\(y\)[/tex]:
[tex]\[ \frac{\partial f}{\partial y} = -\sqrt{3} x + 8y \][/tex]
2. Evaluate the partial derivatives at the given point [tex]\( (\sqrt{3}, 8) \)[/tex]:
[tex]\[ \left. \frac{\partial f}{\partial x} \right|_{(\sqrt{3}, 8)} = 2(\sqrt{3}) - \sqrt{3} \cdot 8 = 2\sqrt{3} - 8\sqrt{3} = -6\sqrt{3} \][/tex]
[tex]\[ \left. \frac{\partial f}{\partial y} \right|_{(\sqrt{3}, 8)} = -\sqrt{3} \cdot \sqrt{3} + 8 \cdot 8 = -3 + 64 = 61 \][/tex]
3. Calculate the slope of the tangent line at the point [tex]\( (\sqrt{3}, 8) \)[/tex]:
The slope [tex]\( m_{\text{tangent}} \)[/tex] of the tangent line is given by:
[tex]\[ m_{\text{tangent}} = -\frac{\left. \frac{\partial f}{\partial x} \right|_{(\sqrt{3}, 8)}}{\left. \frac{\partial f}{\partial y} \right|_{(\sqrt{3}, 8)}} = -\frac{-6\sqrt{3}}{61} = \frac{6\sqrt{3}}{61} \][/tex]
4. Form the equation of the tangent line:
The tangent line passes through the point [tex]\( (\sqrt{3}, 8) \)[/tex] and has the slope [tex]\( \frac{6\sqrt{3}}{61} \)[/tex]. Using the point-slope form of a line:
[tex]\[ y - 8 = \frac{6\sqrt{3}}{61} (x - \sqrt{3}) \][/tex]
So, the equation of the tangent line is:
[tex]\[ y - 8 = \frac{6\sqrt{3}}{61} (x - \sqrt{3}) \][/tex]
5. Calculate the slope of the normal line:
The slope [tex]\( m_{\text{normal}} \)[/tex] of the normal line is the negative reciprocal of the slope of the tangent line:
[tex]\[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{\frac{6\sqrt{3}}{61}} = \frac{61}{6\sqrt{3}} \][/tex]
6. Form the equation of the normal line:
The normal line passes through the point [tex]\( (\sqrt{3}, 8) \)[/tex] and has the slope [tex]\( \frac{61}{6\sqrt{3}} \)[/tex]. Using the point-slope form of a line:
[tex]\[ y - 8 = \frac{61}{6\sqrt{3}} (x - \sqrt{3}) \][/tex]
So, the equation of the normal line is:
[tex]\[ y - 8 = \frac{61}{6\sqrt{3}} (x - \sqrt{3}) \][/tex]
### Final Equations:
- The equation of the tangent line is:
[tex]\[ y - 8 = \frac{6\sqrt{3}}{61} (x - \sqrt{3}) \][/tex]
- The equation of the normal line is:
[tex]\[ y - 8 = \frac{61}{6\sqrt{3}} (x - \sqrt{3}) \][/tex]
So, the normal line is:
[tex]\[ y-8 = \frac{61}{6\sqrt{3}}(x-\sqrt{3}) \][/tex]
