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Sagot :
The slope of the tangent line to the curve at (8, 2) is given by the derivative [tex]\frac{dy}{dx}[/tex] at that point. By the chain rule,
[tex]\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]
Differentiate the given parametric equations with respect to [tex]t[/tex] :
[tex]x = 4t \implies \dfrac{dx}{dt} = 4[/tex]
[tex]y = \dfrac4t \implies \dfrac{dy}{dt} = -\dfrac4{t^2}[/tex]
Then
[tex]\dfrac{dy}{dx} = \dfrac{-\frac4{t^2}}4 = -\dfrac1{t^2}[/tex]
We have [tex]x=8[/tex] and [tex]y=2[/tex] when [tex]t=2[/tex], so the slope at the given point is [tex]\frac{dy}{dx} = -\frac14[/tex].
The normal line to the same point is perpendicular to the tangent line, so its slope is +4. Then using the point-slope formula for a line, the normal line has equation
[tex]y - 2 = 4 (x - 8) \implies \boxed{y = 4x - 30}[/tex]
Alternatively, we can eliminate the parameter and express [tex]y[/tex] explicitly in terms of [tex]x[/tex] :
[tex]x = 4t \implies t = \dfrac x4 \implies y = \dfrac4t = \dfrac4{\frac x4} = \dfrac{16}x[/tex]
Then the slope of the tangent line is
[tex]\dfrac{dy}{dx} = -\dfrac{16}{x^2}[/tex]
At [tex]x = 8[/tex], the slope is again [tex]-\frac{16}{64}=-\frac14[/tex], so the normal has slope +4, and so on.
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