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 $\frac{4a(1+m^2+m^4+1/m^2)}{1+m^2} , \frac{4a(1-m^4)}{m + m^3}$
Best Answer
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 $m$.