Let A, B, and C be the vertices of a variable right-angled triangle inscribed in the parab…

Question

Let $A$, $B$, and $C$ be the vertices of a variable right-angled triangle inscribed in the parabola $y^2=16x$. Let the vertex containing the right angle be $B=(4,8)$, and let the locus of the centroid of $	riangle ABC$ be a conic $\Gamma$. Then three times the length of the latus rectum of $\Gamma$ is \_\_\_\_\_\_\_\_.

Answer: 16

Accepted Answer

Parametrize the parabola as $(4t^2, 8t)$; $B$ corresponds to $t=1$. Let $A=(4t_1^2,8t_1),\ C=(4t_2^2,8t_2)$.

Slope of chord to parameter $t$ from $B(t=1)$ is $\dfrac{2}{1+t}$. Perpendicularity at $B$ gives

$$\frac{2}{1+t_1}\cdot\frac{2}{1+t_2}=-1 \;\Rightarrow\; (1+t_1)(1+t_2)=-4 \;\Rightarrow\; t_1t_2+t_1+t_2=-5.$$

Let $S=t_1+t_2,\ P=t_1t_2$, so $P+S=-5$. The centroid $(h,k)$ of $A,B,C$:

$$h=\frac{4t_1^2+4+4t_2^2}{3}=\frac43(S^2-2P+1),\qquad k=\frac{8t_1+8+8t_2}{3}=\frac83(S+1).$$

Using $P=-5-S$: $h=\dfrac43(S^2+2S+11)$. From $k$: $S=\dfrac{3k}{8}-1$. Substituting and simplifying:

$$h=\frac{3k^2}{16}+\frac{40}{3} \;\Rightarrow\; k^2=\frac{16}{3}\left(h-\frac{40}{3}\right).$$

This is a parabola with $4a=\dfrac{16}{3}$ (latus rectum $=\dfrac{16}{3}$), so three times the latus rectum $=3\times\dfrac{16}{3}=16$.

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