Hello arze Originally Posted by

**arze** $\displaystyle P(ap^2,2ap)$ $\displaystyle Q(aq^2,2aq)$ are on the parabola $\displaystyle y^2=4ax$.

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 $\displaystyle y^2+3a^2=ax$.

Since the chord is a focal chord, pq=-1. and q=-1/p

then $\displaystyle Q(\frac{a}{p^2}, -\frac{2a}{p})$.

the equations of the normals are

$\displaystyle y-2ap=-p(x-ap^2)$_1 and $\displaystyle y+\frac{2a}{p}=\frac{1}{p}(x-\frac{a}{p^2})$_2

1$\displaystyle \rightarrow y=-px+ap^3+2ap$ and

2$\displaystyle \rightarrow y=\frac{x}{p}-\frac{a}{p^3}-\frac{2a}{p}$

solving these two

$\displaystyle \frac{x}{p}-\frac{a}{p^3}-\frac{2a}{p}=-px+ap^3+2ap$

$\displaystyle x=a(t^2+1+\frac{1}{t^2})$

and $\displaystyle y=a(t-\frac{1}{t})$

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:$\displaystyle y=a\left(t-\frac{1}{t}\right)$

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

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

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

Grandad