# Analytic Geometry

• September 28th 2012, 08:16 PM
AuXian
Analytic Geometry
Hello, everybody. I need help... (Worried)
Show that, for all values of p, the point P given by x=ap2, y=2ap lies on the curve y2=4ax.
a) Find the equation of this normal to this curve at the point P.
If this normal meets the curve again at the point Q (aq2, 2aq). Show that p2+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 locus of the point R is y2(x+2a) +4a3=0

I already solved Q (a) and first question of Q (b). However, I can’t solve the last question: “Hence, show that the locus of the point R is y2(x+2a) +4a3=0” in Q (b). Can somebody help me?
Thanks!!(Happy)
• September 28th 2012, 08:29 PM
MarkFL
Re: Analytic Geometry
Presumably, you have found that that $R$ has the coordinates:

$(apq,a(q+p))$

Now, use the relationship between $p$ and $q$ you found earlier:

$p^2+pq+2=0$

which when solved for $q$ is:

$q=-\frac{p^2+2}{p}$

Now you may write the coordinates of $R$ as parametric equations in one parameter, which you may then eliminate to obtain the required Cartesian equation.
• September 28th 2012, 08:34 PM
chiro
Re: Analytic Geometry
Hey AuXian.

I haven't seen this stuff since high school (more than 10 years) but the wiki page gives a derivation of the locus for the parabola:

Parabola - Wikipedia, the free encyclopedia

I'm not sure if you just formulas or have to derive things, but if you just formulas the wiki gives a derivation from start to finish in terms of the a term.