analytics geometry

• Sep 23rd 2012, 07:25 AM
lovelife
analytics geometry
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 .....
• Sep 23rd 2012, 08:50 AM
MarkFL
Re: analytics geometry
How far have you gotten?
• Sep 23rd 2012, 06:23 PM
lovelife
Re: analytics geometry
not yet !
Quote:

Originally Posted by MarkFL2
How far have you gotten?

• Sep 23rd 2012, 07:15 PM
MarkFL
Re: analytics geometry
Have you any ideas about how you might eliminate the parameter p to obtain the Cartesian equation?
• Sep 23rd 2012, 10:50 PM
lovelife
Re: analytics geometry
No...can help me ?
Quote:

Originally Posted by MarkFL2
Have you any ideas about how you might eliminate the parameter p to obtain the Cartesian equation?

• Sep 24th 2012, 06:46 AM
MarkFL
Re: analytics geometry
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 $\displaystyle p$, the point $\displaystyle P(x,y)=\left(ap^2,2ap \right)$ lies on the curve $\displaystyle y^2=4ax$.

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

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

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

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

I will be more than happy to help, but if you have no idea how (or any inclination to try) to eliminate the parameter $\displaystyle p$ 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:

$\displaystyle x=ap^2$

$\displaystyle y=2ap\, \therefore\,p=\frac{y}{2a}$
• Sep 26th 2012, 04:36 AM
lovelife
Re: analytics geometry
Is it the point p substitude into the curve equation ?
• Sep 26th 2012, 06:12 AM
MarkFL
Re: analytics geometry
Where may you substitute for p which will give you a relationship between x and y, thereby eliminating p?
• Sep 26th 2012, 06:48 AM
MarkFL
Re: analytics geometry
Here are two simple examples which may help you see how to eliminate a parameter.

1.) Suppose you are given the parametric equations:

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

$\displaystyle y(t)=7t$

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

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

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

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

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

2.) Suppose you are given the parametric equations:

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

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

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

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

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

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

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

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

$\displaystyle \left(\frac{x}{a} \right)^2+\left(\frac{y}{b} \right)^2=\cos^2(t)+\sin^2(t)$
$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$