# Parabola help!

• May 7th 2010, 04:12 AM
arze
Parabola help!
$P(ap^2,2ap)$ $Q(aq^2,2aq)$ are on the parabola $y^2=4ax$.
If the chord PQ passes through the focus of the parabola and the normals at P and Q meet at R, show that the locus of R is given by $y^2+3a^2=ax$.

Since the chord is a focal chord, pq=-1. and q=-1/p
then $Q(\frac{a}{p^2}, -\frac{2a}{p})$.
the equations of the normals are
$y-2ap=-p(x-ap^2)$_1 and $y+\frac{2a}{p}=\frac{1}{p}(x-\frac{a}{p^2})$_2
1 $\rightarrow y=-px+ap^3+2ap$ and
2 $\rightarrow y=\frac{x}{p}-\frac{a}{p^3}-\frac{2a}{p}$
solving these two
$\frac{x}{p}-\frac{a}{p^3}-\frac{2a}{p}=-px+ap^3+2ap$
$x=a(t^2+1+\frac{1}{t^2})$
and $y=a(t-\frac{1}{t})$
now my problem is I have the parametric equation for the locus of R but i don't know how to get the cartesian equation from here.
Thanks
• May 7th 2010, 05:29 AM
Hello arze
Quote:

Originally Posted by arze
$P(ap^2,2ap)$ $Q(aq^2,2aq)$ are on the parabola $y^2=4ax$.
If the chord PQ passes through the focus of the parabola and the normals at P and Q meet at R, show that the locus of R is given by $y^2+3a^2=ax$.

Since the chord is a focal chord, pq=-1. and q=-1/p
then $Q(\frac{a}{p^2}, -\frac{2a}{p})$.
the equations of the normals are
$y-2ap=-p(x-ap^2)$_1 and $y+\frac{2a}{p}=\frac{1}{p}(x-\frac{a}{p^2})$_2
1 $\rightarrow y=-px+ap^3+2ap$ and
2 $\rightarrow y=\frac{x}{p}-\frac{a}{p^3}-\frac{2a}{p}$
solving these two
$\frac{x}{p}-\frac{a}{p^3}-\frac{2a}{p}=-px+ap^3+2ap$
$x=a(t^2+1+\frac{1}{t^2})$
and $y=a(t-\frac{1}{t})$
now my problem is I have the parametric equation for the locus of R but i don't know how to get the cartesian equation from here.
Thanks

You've done all the hard work! All you need to do now is to say:
$y=a\left(t-\frac{1}{t}\right)$

$\Rightarrow y^2 = a^2\left(t^2-2+\frac{1}{t^2}\right)$
$=a^2\left(t^2+1+\frac{1}{t^2}-3\right)$

$=a^2\left(\frac{x}{a}-3\right)$

$=a^2x-3a^2$