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
`(dy)/(dx)` at that point. By the chain rule,

`(dy)/(dx) = (dy)/(dt) * (dt)/(dx) = ((dy)/(dt))/((dx)/(dt))`

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

`x = 4t \implies (dx)/(dt) = 4`

`y = \frac4t \implies (dy)/(dt) = -\frac4{t^2}`

Then

`(dy)/(dx) = \frac{-\frac4{t^2}}4 = -\frac1{t^2}`

We have
`x=8` and
`y=2` when
`t=2`, so the slope at the given point is
`(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 = \frac x4 \implies y = \frac4t = \frac4{\frac x4} = \frac{16}x`

Then the slope of the tangent line is

`(dy)/(dx) = -(16)/(x^2)`

At
`x = 8`, the slope is again
`-(16)/(64)=-\frac14`, so the normal has slope +4, and so on.