analytics geometry

**lovelife** · September 23rd 2012, 07:25 AM

can help me solve this question ? thanks much !

show that ,for all values of p ,the point p given by x=ap^2 ,y=2ap lies on the curve y^2=4ax .
a)find the equation of the ormal to this curve at the point p.
If this normal meets the curve at the point Q (q^2,2aq) , show that p^2 +pq+2=0 .
b)determine the coordinates of R ,the point of intersection of the tangents of the curve at the point p and Q .
hence ,show that the line locus of the pint R is y^2(x+2a)+4a^3=0 .

thanks .....

**MarkFL** · September 23rd 2012, 08:50 AM

How far have you gotten?

**lovelife** · September 23rd 2012, 06:23 PM

not yet !

Originally Posted by MarkFL2

How far have you gotten?

**MarkFL** · September 23rd 2012, 07:15 PM

Have you any ideas about how you might eliminate the parameter p to obtain the Cartesian equation?

**lovelife** · September 23rd 2012, 10:50 PM

No...can help me ?

Originally Posted by MarkFL2

Have you any ideas about how you might eliminate the parameter p to obtain the Cartesian equation?

**MarkFL** · September 24th 2012, 06:46 AM

Okay, I have worked the problem, but you MUST show some effort here before we can proceed. Also, I found a few typos in the problem statement. Here is how I feel the problem could be stated:

Show that, for all values of $p$ , the point $P(x,y)=\left(ap^2,2ap \right)$ lies on the curve $y^2=4ax$ .

a) Find the equation of the line normal to the curve at $P$ .

If this normal line also crosses the curve at $Q(x,y)=\left(aq^2,2aq \right)$ , show that $p^2+pq+2=0$ .

b) Determine the coordinates of $R$ ,the point of intersection of the tangents of the curve at the points $P$ and $Q$ .

Hence ,show that the line locus of the point $R$ is $y^2(x+2a)+4a^2=0$ .

I will be more than happy to help, but if you have no idea how (or any inclination to try) to eliminate the parameter $p$ to obtain the Cartesian equation, then I am really at a loss to help.

I will give you a nudge to begin. We have the parametric equations:

$x=ap^2$

$y=2ap\, \therefore\,p=\frac{y}{2a}$

**lovelife** · September 26th 2012, 04:36 AM

Is it the point p substitude into the curve equation ?

**MarkFL** · September 26th 2012, 06:12 AM

Where may you substitute for p which will give you a relationship between x and y, thereby eliminating p?

**MarkFL** · September 26th 2012, 06:48 AM

Here are two simple examples which may help you see how to eliminate a parameter.

1.) Suppose you are given the parametric equations:

$x(t)=2t^2+3t+5$

$y(t)=7t$

Notice the equation for $y$ is linear, so we may easily solve for the parameter $t$ :

$t=\frac{y}{7}$

Now, if we substitute for $t$ into the equation for $x$ we get an equation that relates $x$ and $y$ where $t$ has been "eliminated." We then have a Cartesian equation:

$x=2\left(\frac{y}{7} \right)^2+3\left(\frac{y}{7} \right)+5$

$x=\frac{2}{49}y^2+\frac{3}{7}y+5$

2.) Suppose you are given the parametric equations:

$x(t)=a\cdot\cos(t)$

$y(t)=b\cdot\sin(t)$

Rather than solving one of the equations for $t$ , we may rewrite the equations as:

$\frac{x}{a}=\cos(t)$

$\frac{y}{b}=\sin(t)$

If we square both equations, we may then add and take advantage of the Pythagorean identity $\sin^2(\theta)+\cos^2(\theta)=1$

$\left(\frac{x}{a} \right)^2=\cos^2(t)$

$\left(\frac{y}{b} \right)^2=\sin^2(t)$

Adding, we find:

$\left(\frac{x}{a} \right)^2+\left(\frac{y}{b} \right)^2=\cos^2(t)+\sin^2(t)$

Which we may write as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

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