A hyperbola with equation `(x-1)^2/16-(y+2)^2/9=1` is given a triangle is formed with two vertices as the focus of the hyperbola and the third vertix lies on hyperbola . the locus of centroid of triangle is

Options

A. `16(x-1)^2-9(y+2)^2=16`
B. `9(x-1)^2-16(y+2)^2=16`
C. `9(x-1)^2+16(y+2)^2=16`
D. `16(x-1)^2+9(y+2)^2=16`

Answer

To find the locus of the centroid of a triangle formed by the foci of the hyperbola and a point on the hyperbola, we can follow these steps:

### Step 1: Identify the hyperbola and its parameters
The given hyperbola is:
\[ \frac{(x-1)^2}{16} - \frac{(y+2)^2}{9} = 1 \]
From this, we can identify \(a^2 = 16\) and \(b^2 = 9\). Therefore, \(a = 4\) and \(b = 3\).

### Step 2: Find the foci of the hyperbola
The foci of a hyperbola are given by the formula \((\pm ae, 0)\), where \(e\) is the eccentricity. The eccentricity \(e\) can be calculated using:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \]
Thus, the foci are:
\[ F_1 = (5, -2) \quad \text{and} \quad F_2 = (-5, -2) \]

### Step 3: Parametric equations for points on the hyperbola
The parametric equations for points on the hyperbola can be expressed as:
\[ x = 1 + 4 \sec \theta \quad \text{and} \quad y = -2 + 3 \tan \theta \]

### Step 4: Determine the coordinates of the third vertex
The third vertex \(P\) of the triangle, which lies on the hyperbola, has coordinates:
\[ P = (1 + 4 \sec \theta, -2 + 3 \tan \theta) \]

### Step 5: Calculate the centroid of the triangle
The centroid \(G\) of the triangle formed by the foci and the point on the hyperbola is given by:
\[ G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
Substituting the coordinates of the foci and the point \(P\):
\[ G = \left(\frac{5 + (-5) + (1 + 4 \sec \theta)}{3}, \frac{-2 + (-2) + (-2 + 3 \tan \theta)}{3}\right) \]
This simplifies to:
\[ G = \left(\frac{1 + 4 \sec \theta}{3}, \frac{-6 + 3 \tan \theta}{3}\right) \]
\[ G = \left(\frac{1 + 4 \sec \theta}{3}, -2 + \tan \theta\right) \]

### Step 6: Find the locus of the centroid
Let \(G = (h, k)\). From the equations:
\[ h = \frac{1 + 4 \sec \theta}{3} \quad \text{and} \quad k = -2 + \tan \theta \]
We can express \(\sec \theta\) and \(\tan \theta\) in terms of \(h\) and \(k\):
\[ 4 \sec \theta = 3h - 1 \quad \Rightarrow \quad \sec \theta = \frac{3h - 1}{4} \]
\[ \tan \theta = k + 2 \]
Using the identity \(\sec^2 \theta - \tan^2 \theta = 1\):
\[ \left(\frac{3h - 1}{4}\right)^2 - (k + 2)^2 = 1 \]
Expanding and simplifying:
\[ \frac{(3h - 1)^2}{16} - (k + 2)^2 = 1 \]
Multiplying through by 16 gives:
\[ (3h - 1)^2 - 16(k + 2)^2 = 16 \]

### Final Locus Equation
Thus, the locus of the centroid of the triangle is:
\[ (3h - 1)^2 - 16(k + 2)^2 = 16 \]