# Thread: Finding Intersection of Parametric Equation and Axis

1. ## Finding Intersection of Parametric Equation and Axis

Hi, I've been trying to learn the C4 paper ahead of schedule and have gotten down to one last question I can't quite get.

Q5) A curve C has parametric equations $x = at^2$ , $y = 2at$. Show that the equation of the normal to C at the point P, whose parameter is $p$, is:

$px + y - 2ap - ap^3 = 0$

The normal to C at P meets the x-axis at Q. The perpendicular from P to the x-axis meets the x-axis at R. Find the length of QR.

This is my working so far:

$dx/dt = 2at$
$dy/dt = 2a$

$dy/dx = dy/dt * dt/dx = 1/t$ (Chain Rule)

Normal Gradient = $-1/m$ = $-p$

(Use p as parameter for normal equation)

$y - y1 = m(x - x1)$
$y - 2ap = -p(x - ap^2)$
$y - 2ap = -px + ap^3$
$px + y - 2ap - ap^3 = 0$ (As required)

And thats about as far as I got, I've been trying to let y = 0 and let q be the parameter, so:

$qx - 2aq - aq^3 = 0$

But do you then make them equal to each other and figure it out from there? I don't really know, its a WJEC paper too so I can't get any marking schemes without paying, so I thought this way would be better to get an understanding.

Thanks

2. Originally Posted by steve989
Hi, I've been trying to learn the C4 paper ahead of schedule and have gotten down to one last question I can't quite get.

Q5) A curve C has parametric equations $x = at^2$ , $y = 2at$. Show that the equation of the normal to C at the point P, whose parameter is $p$, is:

$px + y - 2ap - ap^3 = 0$

The normal to C at P meets the x-axis at Q. The perpendicular from P to the x-axis meets the x-axis at R. Find the length of QR.

This is my working so far:

$dx/dt = 2at$

$dy/dt = 2a$

$dy/dx = dy/dt * dt/dx = 1/t$ (Chain Rule)

Normal Gradient = $-1/m$ = $-p$

(Use p as parameter for normal equation)

$y - y1 = m(x - x1)$

$y - 2ap = -p(x - ap^2)$

$y - 2ap = -px + ap^3$

$px + y - 2ap - ap^3 = 0$ (As required)

And thats about as far as I got, I've been trying to let y = 0 and let q be the parameter, so:

$qx - 2aq - aq^3 = 0$

But do you then make them equal to each other and figure it out from there? I don't really know, its a WJEC paper too so I can't get any marking schemes without paying, so I thought this way would be better to get an understanding.

Thanks

So you do the hard part and get stuck in an almost trivial part? Putting $y=0$ in the line you get $x=2a+ap^2\Longrightarrow Q=(2a+ap^2,0)$ , and the perpendicular from the point P to the x-axis meets the axis at $(ap^2,0)$ , so...

Tonio

3. Oh yeah, haha that put me in my place, thanks a lot, was a long day, spent 4-5 hours researching unknown topics

Thanks again