1. ## Parametric Equations

The parametric equations of curve C are

x=1/t

y=t²

a) Show that the tangent to C at the point P with parameter p has equation

y+2p³x-3p²=0

b) The tangent to the point C at the point P intersects the x-axis at A and the y-axis at B. Show that PB=2PA

I don't know how to do this question. I know how to find dy/dx but not sure if i need it never attempted one of these questions help pleaase =]

2. ## Parametric equations

Hello djmccabie
Originally Posted by djmccabie
The parametric equations of curve C are

x=1/t

y=t²

a) Show that the tangent to C at the point P with parameter p has equation

y+2p³x-3p²=0

b) The tangent to the point C at the point P intersects the x-axis at A and the y-axis at B. Show that PB=2PA

I don't know how to do this question. I know how to find dy/dx but not sure if i need it never attempted one of these questions help pleaase =]
(a) $x = \frac{1}{t}\Rightarrow \frac{dx}{dt} = -\frac{1}{t^2}$

$y = t^2\Rightarrow \frac{dy}{dt}= 2t$

$\Rightarrow \frac{dy}{dx} = 2t \times (-t^2) = -2t^3$

The point P has parameter $p$; in other words, at P $t = p$. So at P, $x = \frac{1}{p}, y = p^2,\frac{dy}{dx}= -2p^3$

So the tangent at P has equation

$y - p^2 = -2p^3(x - \frac{1}{p})$

$= -2p^3x + 2p^2$

i.e. $y + 2p^3x - 3p^2 = 0$

(b) This line meets the y-axis at B , where $x = 0$ and $y+0-3p^2=0$

$\Rightarrow y = 3p^2$ at B.

So the increase in $y$ from A to P is $p^2$, and the increase in $y$ from P to B is $2p^2$. So by similar triangles, PB = 2PA.