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Sagot :
Let's start by finding the coordinates of the point [tex]\(P\)[/tex] on the curve where [tex]\(\theta = \frac{\pi}{6}\)[/tex].
Given the parametric equations:
[tex]\[ x = 1 - \cos 2\theta \][/tex]
[tex]\[ y = \sin 2\theta \][/tex]
When [tex]\(\theta = \frac{\pi}{6}\)[/tex]:
[tex]\[ x_P = 1 - \cos\left(2 \cdot \frac{\pi}{6}\right) = 1 - \cos\left(\frac{\pi}{3}\right) \][/tex]
[tex]\[ y_P = \sin\left(2 \cdot \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{3}\right) \][/tex]
We know that:
[tex]\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \][/tex]
[tex]\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \][/tex]
Thus, the coordinates of point [tex]\(P\)[/tex] are:
[tex]\[ x_P = 1 - \frac{1}{2} = \frac{1}{2} \][/tex]
[tex]\[ y_P = \frac{\sqrt{3}}{2} \][/tex]
Next, we will determine the slope of the tangent line at point [tex]\(P\)[/tex]. For that, we need to find the derivatives [tex]\(\frac{dx}{d\theta}\)[/tex] and [tex]\(\frac{dy}{d\theta}\)[/tex].
Given:
[tex]\[ x = 1 - \cos 2\theta \][/tex]
[tex]\[ y = \sin 2\theta \][/tex]
The derivatives are:
[tex]\[ \frac{dx}{d\theta} = \frac{d}{d\theta}(1 - \cos 2\theta) = 2 \sin 2\theta \][/tex]
[tex]\[ \frac{dy}{d\theta} = \frac{d}{d\theta}(\sin 2\theta) = 2 \cos 2\theta \][/tex]
At [tex]\(\theta = \frac{\pi}{6}\)[/tex]:
[tex]\[ \frac{dx}{d\theta} \bigg|_{\theta = \frac{\pi}{6}} = 2 \sin\left(2 \cdot \frac{\pi}{6}\right) = 2 \sin\left(\frac{\pi}{3}\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \][/tex]
[tex]\[ \frac{dy}{d\theta} \bigg|_{\theta = \frac{\pi}{6}} = 2 \cos\left(2 \cdot \frac{\pi}{6}\right) = 2 \cos\left(\frac{\pi}{3}\right) = 2 \cdot \frac{1}{2} = 1 \][/tex]
The slope of the tangent line at [tex]\(P\)[/tex] is given by:
[tex]\[ \text{slope of tangent} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{1}{\sqrt{3}} \][/tex]
The slope of the normal line is the negative reciprocal of the slope of the tangent line:
[tex]\[ \text{slope of normal} = -\frac{1}{\frac{1}{\sqrt{3}}} = -\sqrt{3} \][/tex]
We can now write the equation of the normal line in point-slope form:
[tex]\[ y - y_P = \text{slope of normal} \times (x - x_P) \][/tex]
[tex]\[ y - \frac{\sqrt{3}}{2} = -\sqrt{3}(x - \frac{1}{2}) \][/tex]
Rearranging to get it in the form of [tex]\(y = mx + b\)[/tex]:
[tex]\[ y - \frac{\sqrt{3}}{2} = -\sqrt{3}x + \frac{\sqrt{3}}{2} \][/tex]
[tex]\[ y = -\sqrt{3}x + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \][/tex]
[tex]\[ y = -\sqrt{3}x + \sqrt{3} \][/tex]
Adding [tex]\(\sqrt{3}x\)[/tex] to both sides to get it in the desired format:
[tex]\[ y + \sqrt{3}x = \sqrt{3} \][/tex]
Therefore, the equation of the normal to the curve at [tex]\(P\)[/tex] is:
[tex]\[ y + \sqrt{3}x = \sqrt{3} \][/tex]
Given the parametric equations:
[tex]\[ x = 1 - \cos 2\theta \][/tex]
[tex]\[ y = \sin 2\theta \][/tex]
When [tex]\(\theta = \frac{\pi}{6}\)[/tex]:
[tex]\[ x_P = 1 - \cos\left(2 \cdot \frac{\pi}{6}\right) = 1 - \cos\left(\frac{\pi}{3}\right) \][/tex]
[tex]\[ y_P = \sin\left(2 \cdot \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{3}\right) \][/tex]
We know that:
[tex]\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \][/tex]
[tex]\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \][/tex]
Thus, the coordinates of point [tex]\(P\)[/tex] are:
[tex]\[ x_P = 1 - \frac{1}{2} = \frac{1}{2} \][/tex]
[tex]\[ y_P = \frac{\sqrt{3}}{2} \][/tex]
Next, we will determine the slope of the tangent line at point [tex]\(P\)[/tex]. For that, we need to find the derivatives [tex]\(\frac{dx}{d\theta}\)[/tex] and [tex]\(\frac{dy}{d\theta}\)[/tex].
Given:
[tex]\[ x = 1 - \cos 2\theta \][/tex]
[tex]\[ y = \sin 2\theta \][/tex]
The derivatives are:
[tex]\[ \frac{dx}{d\theta} = \frac{d}{d\theta}(1 - \cos 2\theta) = 2 \sin 2\theta \][/tex]
[tex]\[ \frac{dy}{d\theta} = \frac{d}{d\theta}(\sin 2\theta) = 2 \cos 2\theta \][/tex]
At [tex]\(\theta = \frac{\pi}{6}\)[/tex]:
[tex]\[ \frac{dx}{d\theta} \bigg|_{\theta = \frac{\pi}{6}} = 2 \sin\left(2 \cdot \frac{\pi}{6}\right) = 2 \sin\left(\frac{\pi}{3}\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \][/tex]
[tex]\[ \frac{dy}{d\theta} \bigg|_{\theta = \frac{\pi}{6}} = 2 \cos\left(2 \cdot \frac{\pi}{6}\right) = 2 \cos\left(\frac{\pi}{3}\right) = 2 \cdot \frac{1}{2} = 1 \][/tex]
The slope of the tangent line at [tex]\(P\)[/tex] is given by:
[tex]\[ \text{slope of tangent} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{1}{\sqrt{3}} \][/tex]
The slope of the normal line is the negative reciprocal of the slope of the tangent line:
[tex]\[ \text{slope of normal} = -\frac{1}{\frac{1}{\sqrt{3}}} = -\sqrt{3} \][/tex]
We can now write the equation of the normal line in point-slope form:
[tex]\[ y - y_P = \text{slope of normal} \times (x - x_P) \][/tex]
[tex]\[ y - \frac{\sqrt{3}}{2} = -\sqrt{3}(x - \frac{1}{2}) \][/tex]
Rearranging to get it in the form of [tex]\(y = mx + b\)[/tex]:
[tex]\[ y - \frac{\sqrt{3}}{2} = -\sqrt{3}x + \frac{\sqrt{3}}{2} \][/tex]
[tex]\[ y = -\sqrt{3}x + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \][/tex]
[tex]\[ y = -\sqrt{3}x + \sqrt{3} \][/tex]
Adding [tex]\(\sqrt{3}x\)[/tex] to both sides to get it in the desired format:
[tex]\[ y + \sqrt{3}x = \sqrt{3} \][/tex]
Therefore, the equation of the normal to the curve at [tex]\(P\)[/tex] is:
[tex]\[ y + \sqrt{3}x = \sqrt{3} \][/tex]
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