# Locus of vertex of a rectangle

If from the vertex of a parabola $y^2 = 4ax$ a pair of chords be drawn at right angles to one another and with these chords as adjacent sides a rectangle be constructed , then we have to find the locus of the outer corner of the rectangle .

I tried ,

Let the equation of one chord be $y = mx$

And meet the parabola at $(4a/m^2 , 4a/m)$

Other chord is $y = -x/m$

Meet at $(4am^2 , -4am)$

Then wrote the equation of lines perpendicular to the chords through their respective points on the parabola .

After solving got the point of intersection . But not able to find the locus .

The point is $\frac{4a(1+m^2+m^4+1/m^2)}{1+m^2} , \frac{4a(1-m^4)}{m + m^3}$

• You say you "got the point of intersection." I assume that you have a point whose coordinates are functions of $m$. What expression did you get? Also, what do you mean by "find the locus"? Do you want a parameterization of the curve, a Cartesian equation, or a geometric description? Your "point of intersection" would be a parameterization! – Rory Daulton Dec 30 '15 at 12:44
• @RoryDaulton the point is there in edited question – user101522 Dec 30 '15 at 12:49
• Thanks. But what do you mean by "find the locus"? You now have a parameterization. – Rory Daulton Dec 30 '15 at 13:00
• Locus means that we have to find a equation for which the point satisfy – user101522 Dec 30 '15 at 13:03
• Does the chord have to pass through the origin? – KittyL Dec 30 '15 at 13:16

First, the coordinates of the outside corner point can be simply written as $$(\frac{4a}{m^2}+4am^2, \frac{4a}{m}-4am)$$ by using vector addition. Yours is correct too. It can be simplified to the same value.
Then square the $y$ value gives you that $y^2=16a^2(\frac{1}{m^2}+m^2-2)$. Now notice that the $x$ value contains a factor $\frac{1}{m^2}+m^2$. You can then do substitution to remove the $m$.