If `r_(1), r_(2)` be the length of the perpendicular chords of the parabola `y^(2) = 4ax` drawn through the vertex, then show that `(r_(1)r_(2))^(4//3) = 16a^(2)(r_(1)^(2//3) +r_(2)^(2//3))`

Explanation

Since chord are perpendicular, therefore if one makes an angle `(90^(@)-theta)` then the other will make an angle `(90^(@) – theta)` with x-axis
Let `AP=r_(1)` and `AQ=r_(2)`
If `anglePAX=theta` then `angleQAX=90^(@)-theta`
`theta` Co-ordinates of P and Q are (`r_(1) costheta, r_(1) sintheta`)
and `(r_(2)sintheta,-r_(2)costheta)` respectively.

Since P and Q lies on `y^(2) = 4ax`
`thereforer_(1)^(2)sin^(2)theta=4ar_(1)costheta`
and `r_(2)^(2)cos^(2)theta=4ar_(2)sintheta`
`!r_(1)=(4acostheta)/(sin^(2)theta)`
and `r_(2)=(4asintheta)/(cos^(2)theta)`
`therefore(r_(1)r_(2))^(4//3)`
`=((4acostheta)/(sin^(2)theta).(4asintheta)/(cos^(2)theta))^(4//3)`
`=((16a^(2))/(sinthetacostheta))^(4//3)` ...(i)
and `16a^(2).(r_(1)^(2//3)+r_(2)^(2//3))`
`=16a^(2){((4acostheta)/(sin^(2)theta))^(2//3)+((4asintheta)//(cos^(2)theta))^(2//3)}`
`=16a^(2).(4a)^(2//3){((costheta)^(2//3))/((sintheta)^(4//3))+((sintheta)^(2//3))/((costheta)^(4//3))}`
`=16a^(2).(4a)^(2//3){(cos^(2)theta+sin^(2)theta)/((sintheta)^(4//3))+((sintheta)^(2//3))/((costheta)^(4//3))}`
`=(16a^(3).(4a)^(2//3))/((sinthetacostheta)^(4//3))`
`=((16a^(2))/(sinthetacostheta))^(4//3)`
`=(r_(1)r_(2))^(4//3)` [from (i)]