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(a) Find the equation of the normal to the hyperbola x = 4t, y = 4/t at the point (8,2).

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.