Finding Intersection of Parametric Equation and Axis

**steve989** · March 31st 2010, 02:41 AM

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

**tonio** · March 31st 2010, 04:13 AM

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

**steve989** · April 2nd 2010, 05:01 AM

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

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