Question

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

Answer 1

The slope of the tangent line to the curve at (8, 2) is given by the derivative $\frac{dy}{dx}$ at that point. By the chain rule,

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Differentiate the given parametric equations with respect to $t$ :

$x = 4t \implies \dfrac{dx}{dt} = 4$

$y = \dfrac4t \implies \dfrac{dy}{dt} = -\dfrac4{t^2}$

Then

$\dfrac{dy}{dx} = \dfrac{-\frac4{t^2}}4 = -\dfrac1{t^2}$

We have $x=8$ and $y=2$ when $t=2$, so the slope at the given point is $\frac{dy}{dx} = -\frac14$.

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

$y - 2 = 4 (x - 8) \implies \boxed{y = 4x - 30}$

Alternatively, we can eliminate the parameter and express $y$ explicitly in terms of $x$ :

$x = 4t \implies t = \dfrac x4 \implies y = \dfrac4t = \dfrac4{\frac x4} = \dfrac{16}x$

Then the slope of the tangent line is

$\dfrac{dy}{dx} = -\dfrac{16}{x^2}$

At $x = 8$, the slope is again $-\frac{16}{64}=-\frac14$, so the normal has slope +4, and so on.