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Sagot :
Applying derivative concepts, the y-intercept of the line tangent to the curve at (2√3, √3) is: [tex]y = \sqrt{3}[/tex]
The equation of the line is:
[tex]y - y_1 = m(x - x_1)[/tex]
In which:
- There is a point [tex](x_1, y_1)[/tex].
- The slope is [tex]m = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}[/tex]
In this problem, point [tex](2\sqrt{3}, \sqrt{3})[/tex], hence [tex]x_1 = 2\sqrt{3}, y_1 = \sqrt{3}[/tex]
To find the value of [tex]\theta[/tex], we have that:
[tex]3\tan{\theta} = \sqrt{3}[/tex]
[tex]\tan{\theta} = \frac{\sqrt{3}}{3}[/tex]
[tex]\theta = \frac{\pi}{6}[/tex]
Then, for the slope:
[tex]\frac{dy}{d\theta} = 3\sec^{2}{\theta}[/tex]
[tex]\frac{dx}{d\theta} = 3\sec{\theta}\tan{\theta}[/tex]
Considering [tex]\theta = \frac{\pi}{6}[/tex]:
[tex]\frac{dy}{d\theta} = 3\left(\frac{2}{\sqrt{3}}\right)^2 = 2[/tex]
[tex]\frac{dx}{d\theta} = 3\left(\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{3}\right) = 2[/tex]
[tex]m = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = 1[/tex]
Then, the equation of the line is:
[tex]y - y_1 = m(x - x_1)[/tex]
[tex]y - \sqrt{3} = x - 2\sqrt{3}[/tex]
[tex]y = x - \sqrt{3}[/tex]
The y-intercept is the value of y when x = 0, hence:
[tex]y = \sqrt{3}[/tex]
A similar problem is given at https://brainly.com/question/22426360
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