Question

A curve is given by the parametric equations x=t,y=`1/t`. (pls see details below). ?

(a) Find the equation of line passing through the points (s,`1/s`) and (t,`1/t`).
(b) Deduce the equation of tangent to the curve at the point (t,`1/t`).
(c)Find the equation of normal to the curve at the point (t,`1/t`).
If this normal cuts the curve again, find the coordinates of point of intersection

Answer 1

Line through...: `tx + t^2 s y = t^2-s^2`

Tangent: `x +t^2 y=2t`

Normal: `t^3 x - t y = t^4 -1`

Other intersection: `(-1/t^3,-t^3)`

Explanation:

Fun. We have `(x,y)=(t,1/t).` To deparameterize (not that we have to) that's `x=t` so this is just the standard hyperbola `y=1/x.`

The line through `(s.1/s)` and `(t,1/t)` is

`(y-1/s)(t-s)=(x-s)(1/t - 1/s)`

Multiplying both sides by `st`,

`t(sy-1)(t-s)=(x-s)(s-t)`

`t(sy-1)=-(x-s)`

`x + sty = s+t`

We get the tangent line as `s to t`

`x +t^2 y=2t`

For the normal line we swap the coefficients on `x` and `y`, negating one, then use the point `(t,1/t)` for the constant:

`t^2 x - y = t^2(t)-1/t = t^3 -1/t`

`t^3 x - t y = t^4 -1`

We plug in `y=1/x` to find where the normal meets the curve

`t^3 x - t/x = t^4 -1`

Multiply by `x` to clear the denominator.

`t^3 x^2 - (t^4-1)x - t = 0`

We know `x=t` is one root so we can factor:

`(x-t)(t^3 x + 1) = 0`

`x=t` we know about; the other point is `(-1/t^3,-t^3)`

Let's plot these at `t=2` so `(x,y)=(2,1/2)` with the Socratic `cancel{"crasher"}` grapher.

Plot `0=(y-1/x)(x +2^2 y - 2(2))(2^3 x - 2 y - (2^4 -1))( (x- (-1/2^3) )^2+(y - (-2)^3)^2-.1^2)`

graph{0=(y-1/x)(x +2^2 y - 2(2))(2^3 x - 2 y - (2^4 -1))( (x- (-1/2^3) )^2+(y - (-2)^3)^2-.2^2) [-20, 20, -10, 10]}