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Sagot :
To find the tangent and normal lines to the graph of [tex]\(x^2 - \sqrt{3}xy + 4y^2 = 235\)[/tex] at the point [tex]\((\sqrt{3}, 8)\)[/tex], follow these steps:
1. Find Partial Derivatives:
Compute the partial derivatives of the given equation with respect to [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
[tex]\[ F(x, y) = x^2 - \sqrt{3}xy + 4y^2 - 235 \][/tex]
The partial derivative with respect to [tex]\(x\)[/tex] is:
[tex]\[ F_x = \frac{\partial F}{\partial x} = 2x - \sqrt{3}y \][/tex]
The partial derivative with respect to [tex]\(y\)[/tex] is:
[tex]\[ F_y = \frac{\partial F}{\partial y} = -\sqrt{3}x + 8y \][/tex]
2. Evaluate Partial Derivatives at Given Point:
Evaluate [tex]\(F_x\)[/tex] and [tex]\(F_y\)[/tex] at the point [tex]\((\sqrt{3}, 8)\)[/tex].
[tex]\[ F_x(\sqrt{3}, 8) = 2(\sqrt{3}) - \sqrt{3}(8) = 2\sqrt{3} - 8\sqrt{3} = -6\sqrt{3} \][/tex]
[tex]\[ F_y(\sqrt{3}, 8) = -\sqrt{3}(\sqrt{3}) + 8(8) = -3 + 64 = 61 \][/tex]
3. Find the Slope of the Tangent Line:
The slope of the tangent line to the curve at [tex]\((\sqrt{3}, 8)\)[/tex] is given by:
[tex]\[ \text{slope of tangent line} = -\frac{F_x}{F_y} = -\frac{-6\sqrt{3}}{61} = \frac{6\sqrt{3}}{61} \][/tex]
4. Find the Equation of the Tangent Line:
The equation of the tangent line at [tex]\((\sqrt{3}, 8)\)[/tex] is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Substituting [tex]\(m = \frac{6\sqrt{3}}{61}\)[/tex], [tex]\(x_1 = \sqrt{3}\)[/tex], and [tex]\(y_1 = 8\)[/tex]:
[tex]\[ y - 8 = \frac{6\sqrt{3}}{61}(x - \sqrt{3}) \][/tex]
5. Find the Slope of the Normal Line:
The slope of the normal line is the negative reciprocal of the slope of the tangent line:
[tex]\[ \text{slope of normal line} = -\frac{1}{\text{slope of tangent line}} = -\frac{1}{\frac{6\sqrt{3}}{61}} = -\frac{61}{6\sqrt{3}} = -\frac{61\sqrt{3}}{18} \][/tex]
6. Find the Equation of the Normal Line:
The equation of the normal line at [tex]\((\sqrt{3}, 8)\)[/tex] is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Substituting [tex]\(m = -\frac{61\sqrt{3}}{18}\)[/tex], [tex]\(x_1 = \sqrt{3}\)[/tex], and [tex]\(y_1 = 8\)[/tex]:
[tex]\[ y - 8 = -\frac{61\sqrt{3}}{18}(x - \sqrt{3}) \][/tex]
Thus, the equations are:
- The tangent line is [tex]\[ y - 8 = \frac{6\sqrt{3}}{61}(x - \sqrt{3}) \][/tex]
- The normal line is [tex]\[ y - 8 = -\frac{61\sqrt{3}}{18}(x - \sqrt{3}) \][/tex]
1. Find Partial Derivatives:
Compute the partial derivatives of the given equation with respect to [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
[tex]\[ F(x, y) = x^2 - \sqrt{3}xy + 4y^2 - 235 \][/tex]
The partial derivative with respect to [tex]\(x\)[/tex] is:
[tex]\[ F_x = \frac{\partial F}{\partial x} = 2x - \sqrt{3}y \][/tex]
The partial derivative with respect to [tex]\(y\)[/tex] is:
[tex]\[ F_y = \frac{\partial F}{\partial y} = -\sqrt{3}x + 8y \][/tex]
2. Evaluate Partial Derivatives at Given Point:
Evaluate [tex]\(F_x\)[/tex] and [tex]\(F_y\)[/tex] at the point [tex]\((\sqrt{3}, 8)\)[/tex].
[tex]\[ F_x(\sqrt{3}, 8) = 2(\sqrt{3}) - \sqrt{3}(8) = 2\sqrt{3} - 8\sqrt{3} = -6\sqrt{3} \][/tex]
[tex]\[ F_y(\sqrt{3}, 8) = -\sqrt{3}(\sqrt{3}) + 8(8) = -3 + 64 = 61 \][/tex]
3. Find the Slope of the Tangent Line:
The slope of the tangent line to the curve at [tex]\((\sqrt{3}, 8)\)[/tex] is given by:
[tex]\[ \text{slope of tangent line} = -\frac{F_x}{F_y} = -\frac{-6\sqrt{3}}{61} = \frac{6\sqrt{3}}{61} \][/tex]
4. Find the Equation of the Tangent Line:
The equation of the tangent line at [tex]\((\sqrt{3}, 8)\)[/tex] is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Substituting [tex]\(m = \frac{6\sqrt{3}}{61}\)[/tex], [tex]\(x_1 = \sqrt{3}\)[/tex], and [tex]\(y_1 = 8\)[/tex]:
[tex]\[ y - 8 = \frac{6\sqrt{3}}{61}(x - \sqrt{3}) \][/tex]
5. Find the Slope of the Normal Line:
The slope of the normal line is the negative reciprocal of the slope of the tangent line:
[tex]\[ \text{slope of normal line} = -\frac{1}{\text{slope of tangent line}} = -\frac{1}{\frac{6\sqrt{3}}{61}} = -\frac{61}{6\sqrt{3}} = -\frac{61\sqrt{3}}{18} \][/tex]
6. Find the Equation of the Normal Line:
The equation of the normal line at [tex]\((\sqrt{3}, 8)\)[/tex] is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Substituting [tex]\(m = -\frac{61\sqrt{3}}{18}\)[/tex], [tex]\(x_1 = \sqrt{3}\)[/tex], and [tex]\(y_1 = 8\)[/tex]:
[tex]\[ y - 8 = -\frac{61\sqrt{3}}{18}(x - \sqrt{3}) \][/tex]
Thus, the equations are:
- The tangent line is [tex]\[ y - 8 = \frac{6\sqrt{3}}{61}(x - \sqrt{3}) \][/tex]
- The normal line is [tex]\[ y - 8 = -\frac{61\sqrt{3}}{18}(x - \sqrt{3}) \][/tex]
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