Conic Sections – Perpendicular Chord in Parabola

Question

If $r_1,r_2$ be the length of the perpendicular chords drawn through the vertex of a parabola $y^2=4ax$, then show that
$$(r_1r_2)^{4/3}=16a^2(r_1^{2/3}+r_2^{2/3})$$

Answer 1

Let $O=(0,0)$ be the vertex of the parabola, and $P_1=(x_1,y_1)$, $P_2=(x_2,y_2)$, the other endpoints of the chords. The perpendicularity condition yields $$ y_1y_2=-x_1x_2=-{1\over16a^2}y_1^2y_2^2, \quad\hbox{that is}\quad y_1y_2=-16a^2 \quad\hbox{and}\quad x_1x_2=16a^2. $$ We have then: $$ r_2^2=x_2^2+y_2^2=x_2^2+4ax_2={256a^4\over x_1^2}+{64a^3\over x_1}= {64a^3\over x_1^3}(4ax_1+x_1^2)={64a^3\over x_1^3}(y_1^2+x_1^2) =\left({4a\over x_1}\right)^3r_1^2 $$ that is $$ r_2^{2/3}={4a\over x_1}r_1^{2/3}. $$ It follows that $$ r_1^{4/3}r_2^{4/3}={16a^2\over x_1^2}r_1^{8/3} $$ and $$ r_1^{2/3}+r_2^{2/3}=\left(1+{4a\over x_1}\right)r_1^{2/3} ={x_1^2+4ax_1\over x_1^2}r_1^{2/3} ={r_1^2\over x_1^2}r_1^{2/3}={1\over x_1^2}r_1^{8/3}. $$ In conclusion: $$ {r_1^{4/3}r_2^{4/3}\over r_1^{2/3}+r_2^{2/3}}=16a^2. $$