Let A,B and C be the vertices of a variable right angled triangle inscribed in the parabola y^2=16x. Let the vertex B containing the right angle be…

The equation of the parabola is $y^2 = 16x$y2=16x, which gives $a = 4$a=4. Any point on the parabola can be taken as $(4t^2, 8t)$(4t2,8t).

The vertex $B$B is given as $(4,8)$(4,8), which corresponds to the parameter $t = 1$t=1. Let the parameters for vertices $A$A and $C$C be $t_1$t1​ and $t_2$t2​ respectively. Thus, $A \equiv (4t_1^2, 8t_1)$A≡(4t12​,8t1​) and $C \equiv (4t_2^2, 8t_2)$C≡(4t22​,8t2​).

The slope of $AB$AB is:

$$m_{AB} = \dfrac{8t_1 - 8}{4t_1^2 - 4} = \dfrac{2(t_1 - 1)}{(t_1 - 1)(t_1 + 1)} = \dfrac{2}{t_1 + 1}$$mAB​=4t12​−48t1​−8​=(t1​−1)(t1​+1)2(t1​−1)​=t1​+12​

Similarly, the slope of $BC$BC is:

$$m_{BC} = \dfrac{2}{t_2 + 1}$$mBC​=t2​+12​

Since $\triangle ABC$△ABC is right-angled at $B$B, $AB \perp BC$AB⊥BC. Therefore, $m_{AB} \times m_{BC} = -1$mAB​×mBC​=−1:

$$\dfrac{2}{t_1 + 1} \times \dfrac{2}{t_2 + 1} = -1$$t1​+12​×t2​+12​=−1

$$(t_1 + 1)(t_2 + 1) = -4$$(t1​+1)(t2​+1)=−4

$$t_1t_2 + t_1 + t_2 + 5 = 0$$t1​t2​+t1​+t2​+5=0

Let the centroid of $\triangle ABC$△ABC be $(h, k)$(h,k). Then:

$$h = \dfrac{4 + 4t_1^2 + 4t_2^2}{3} \Rightarrow t_1^2 + t_2^2 = \dfrac{3h}{4} - 1$$h=34+4t12​+4t22​​⇒t12​+t22​=43h​−1

$$k = \dfrac{8 + 8t_1 + 8t_2}{3} \Rightarrow t_1 + t_2 = \dfrac{3k}{8} - 1$$k=38+8t1​+8t2​​⇒t1​+t2​=83k​−1

From the perpendicularity condition, we can express $t_1t_2$t1​t2​ as:

$$t_1t_2 = -5 - (t_1 + t_2) = -5 - \left(\dfrac{3k}{8} - 1\right) = -4 - \dfrac{3k}{8}$$t1​t2​=−5−(t1​+t2​)=−5−(83k​−1)=−4−83k​

Using the algebraic identity $t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1t_2$t12​+t22​=(t1​+t2​)2−2t1​t2​, we substitute the expressions in terms of $k$k:

$$\dfrac{3h}{4} - 1 = \left(\dfrac{3k}{8} - 1\right)^2 - 2\left(-4 - \dfrac{3k}{8}\right)$$43h​−1=(83k​−1)2−2(−4−83k​)

$$\dfrac{3h}{4} - 1 = \dfrac{9k^2}{64} - \dfrac{3k}{4} + 1 + 8 + \dfrac{3k}{4}$$43h​−1=649k2​−43k​+1+8+43k​

$$\dfrac{3h}{4} - 1 = \dfrac{9k^2}{64} + 9$$43h​−1=649k2​+9

$$\dfrac{3h}{4} = \dfrac{9k^2}{64} + 10$$43h​=649k2​+10

$$3h = \dfrac{9k^2}{16} + 40$$3h=169k2​+40

$$\dfrac{9k^2}{16} = 3h - 40$$169k2​=3h−40

$$k^2 = \dfrac{16}{9}(3h - 40) = \dfrac{16}{3}\left(h - \dfrac{40}{3}\right)$$k2=916​(3h−40)=316​(h−340​)

Replacing $(h, k)$(h,k) with $(x, y)$(x,y), the locus $C_o$Co​ is the parabola:

$$y^2 = \dfrac{16}{3}\left(x - \dfrac{40}{3}\right)$$y2=316​(x−340​)

The length of the latus rectum of this parabola is $\dfrac{16}{3}$316​.

Three times the length of the latus rectum is $3 \times \dfrac{16}{3} = 16$3×316​=16.

Answer: $16$16