# Finding Intersection of Parametric Equation and Axis

• Mar 31st 2010, 02:41 AM
steve989
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 \$\displaystyle x = at^2\$ , \$\displaystyle y = 2at\$. Show that the equation of the normal to C at the point P, whose parameter is \$\displaystyle p\$, is:

\$\displaystyle 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:

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

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

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

(Use p as parameter for normal equation)

\$\displaystyle y - y1 = m(x - x1)\$
\$\displaystyle y - 2ap = -p(x - ap^2)\$
\$\displaystyle y - 2ap = -px + ap^3\$
\$\displaystyle 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:

\$\displaystyle 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

• Mar 31st 2010, 04:13 AM
tonio
Quote:

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 \$\displaystyle x = at^2\$ , \$\displaystyle y = 2at\$. Show that the equation of the normal to C at the point P, whose parameter is \$\displaystyle p\$, is:

\$\displaystyle 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:

\$\displaystyle dx/dt = 2at\$

\$\displaystyle dy/dt = 2a\$

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

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

(Use p as parameter for normal equation)

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

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

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

\$\displaystyle 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:

\$\displaystyle 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? (Giggle) Putting \$\displaystyle y=0\$ in the line you get \$\displaystyle 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 \$\displaystyle (ap^2,0)\$ , so...

Tonio
• Apr 2nd 2010, 05:01 AM
steve989
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