Join the IDNLearn.com community and start getting the answers you need today. Receive prompt and accurate responses to your questions from our community of knowledgeable professionals ready to assist you at any time.
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]
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For clear and precise answers, choose IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.
