Question

The parametric equations of a curve are x=`t^2` and y=`3/t`. Points P and Q with parameters p and q respectively lie on the curve. (pls see below). ?

(a) Find the gradient of chord PQ and deduce the gradient of the tangent to the curve at point P
(b) Find the coordinates of points A and B, when the tangent at P meets the x-axis and y-axis respectively.Show that area of triangle OAB is `9/2`P units.

Answer 1

The slope of the tangent at `(p^2, 3/p)` is `-3/{2p^3}` and the area of the triangle seems to be `27/4 p` not `9/2 p.`

Explanation:

`x=t^2`, `y=3/t`

`y^2=9/t^2=9/x`

`xy^2 = 9`

Third degree; hyperbola-like in the first and fourth quadrants.

We have `P(p^2,3/p)` and `Q(q^2,3/q)`

The slope (aka gradient) between them is

`g(p,q) = (3/q-3/p)/{q^2-p^2} = {3(p-q)}/{pq(q-p)(q+p)} = -3/{pq(p+q)`

The tangent slope is

`m = lim_{q to p} g(p,q) = -3/{p^2(2p)} = -3/{2p^3}`

The line through `P(p^2,3/p)` with that slope is

`y-3/p = -3/{2p^3} (x - p^2)`

`2p^3y - 6p^2 = -3x + 3 p^2`

`3 x + 2 p^3 y = 9 p^2`

We have x intercept `A(3p^2, 0)` and y intercept `(0, 9/(2p))`

This is a right triangle because the sides are the axes. So the legs are the altitudes and bases. The area is

`S = 1/2 3p^2 (9/(2p)) = 27/4 p`

That's different than what the question said. Let's try it for `p=1`,
`P(1,3)`

Tangent line:

`3 x + 2 y = 9`

x intercept `(3,0),` y intercept `(0,9/2)` area

`1/2 (3) (9/2) = 27/4`

I think the question is wrong. I'll graph for `p=1` after I post.

Plot `0=(xy^2-9)(3 x + 2 y - 9)`

graph{ 0=(xy^2-9)(3 x + 2 y - 9) [-4.67, 15.33, -3.84, 6.16]}