Question

Let point A be the point of intersection tangents at points B and C of parabola $y^2=4x$. If A lies on directrix, then the locus of centroid of $\triangle ABC$ is a conic. Find the length of its latus rectum.

Answer 1

3

Answer 2

The given parabola is $y^2=4x$, which corresponds to the standard form $y^2=4ax$ with $a=1$. Let the points B and C on the parabola be $B(t_1^2, 2t_1)$ and $C(t_2^2, 2t_2)$. The point of intersection of tangents at $t_1$ and $t_2$ to the parabola $y^2=4ax$ is given by $A(at_1t_2, a(t_1+t_2))$. For $y^2=4x$ (with $a=1$), point A is $A(t_1t_2, t_1+t_2)$. The directrix of the parabola $y^2=4ax$ is $x=-a$. For $y^2=4x$, the directrix is $x=-1$. Given that point A lies on the directrix, its x-coordinate must be $-1$. Thus, $t_1t_2 = -1$.

Let the centroid of $\triangle ABC$ be $G(h, k)$. The coordinates of the vertices are $A(t_1t_2, t_1+t_2)$, $B(t_1^2, 2t_1)$, and $C(t_2^2, 2t_2)$. The coordinates of the centroid are: $h = \frac{t_1t_2 + t_1^2 + t_2^2}{3}$ $k = \frac{(t_1+t_2) + 2t_1 + 2t_2}{3} = \frac{3(t_1+t_2)}{3} = t_1+t_2$.

We know that $(t_1+t_2)^2 = t_1^2 + t_2^2 + 2t_1t_2$. So, $t_1^2 + t_2^2 = (t_1+t_2)^2 - 2t_1t_2$. Substitute $t_1+t_2 = k$ and $t_1t_2 = -1$: $t_1^2 + t_2^2 = k^2 - 2(-1) = k^2 + 2$.

Now substitute these into the equation for $h$: $h = \frac{(-1) + (k^2+2)}{3} = \frac{k^2+1}{3}$. Rearranging this equation gives: $3h = k^2 + 1$ $k^2 = 3h - 1$.

The locus of the centroid G is $y^2 = 3x - 1$. This can be rewritten as $y^2 = 3(x - 1/3)$. This is the equation of a parabola in the form $Y^2 = 4a'X$, where $Y=y$, $X=x-1/3$, and $4a' = 3$. The length of the latus rectum of a parabola $Y^2 = 4a'X$ is $4a'$. Therefore, the length of the latus rectum of the locus is 3.