Animated Solution for Mathematics - Conic Sections: The locus of the centroid of the triangle formed by any point $P$ on the hyperbola $16x^2-9y^2+32x+36y-164=0$, and its foci is :

Imagine you are standing on a vast, perfectly flat plane. Before you lies a hyperbola, a curve defined by the difference of distances from two fixed points—the foci.

You pick a point $P$ anywhere on this curve and connect it to the two foci, $S_1$ and $S_2$. Suddenly, you have a triangle, $△P{S}_1{S}_2$.

As you slide $P$ along the hyperbola, the triangle morphs, stretches, and shifts. But there is a hidden order in this chaos: the centroid of this triangle, the point where its medians intersect, traces its own path. Today, we are going to find that path.

First, we must understand our foundation. We are given the equation $16x^2-9y^2+32x+36y-164=0$. This looks intimidating, but it is merely a hyperbola in disguise.

To reveal its true form, we use the method of completing the square. We group the $x$ terms and the $y$ terms:

By adding and subtracting the necessary constants inside the parentheses, we transform this into:

This simplifies beautifully to $16(x+1{)}^2-9(y-2{)}^2=144$. Dividing by $144$, we arrive at the standard form:

Now, the hyperbola speaks to us. Its center is at $C(-1,2)$, and we can clearly see $a^2=9$ and $b^2=16$. This is the stage upon which our point $P$ dances.

To define our triangle, we need the foci, $S_1$ and $S_2$. The eccentricity $e$ is our key, calculated as:

The distance from the center to each focus is $ae=3\times \frac{5}{3}=5$. Since the transverse axis is horizontal ($y=2$), the foci are located at $(-1\pm 5,2)$, giving us $S_1(4,2)$ and $S_2(-6,2)$.

These are the fixed vertices of our triangle.

Let $P(\alpha ,\beta )$ be any point on the hyperbola. The centroid $G(h,k)$ of $△P{S}_1{S}_2$ is the average of its vertices:

We need the locus of $G$, so we must express $\alpha$ and $\beta$ in terms of $h$ and $k$. A quick rearrangement gives us $\alpha =3h+2$ and $\beta =3k-4$.

Since $P$ lies on the hyperbola, it must satisfy the equation $\frac{(\alpha +1{)}^2}{9}-\frac{(\beta -2{)}^2}{16}=1$. Substituting our expressions for $\alpha$ and $\beta$:

Factoring out the $3$ from the numerators, we get $\frac{9(h+1{)}^2}{9}-\frac{9(k-2{)}^2}{16}=1$, which simplifies to $(h+1{)}^2-\frac{9(k-2{)}^2}{16}=1$. Multiplying by $16$ and expanding, we finally obtain:

Replacing $h$ and $k$ with $x$ and $y$, we find the locus: $16x^2-9y^2+32x+36y-36=0$. The centroid traces a path that is itself a hyperbola, a perfect, scaled reflection of the original. You have successfully navigated the geometry of motion!