Find the locus of the foot of perpendicular from the centre upon any normal to line hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`.

Answer

To find the locus of the foot of the perpendicular from the center of the hyperbola to any normal, we will follow these steps:

### Step 1: Write the equation of the hyperbola
The given hyperbola is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 2: Write the equation of the normal to the hyperbola
The equation of the normal to the hyperbola at a point \((x_0, y_0)\) is given by:
\[ \frac{Ax}{\sec \theta} + \frac{By}{\tan \theta} = A^2 + B^2 \]
For our hyperbola, this becomes:
\[ \frac{ax_0}{\sec \theta} + \frac{by_0}{\tan \theta} = a^2 + b^2 \]
Let’s denote this equation as (1).

### Step 3: Calculate the slope of the normal
The slope \(m\) of the normal can be derived from the equation:
\[ m = -\frac{b}{a} \tan \theta \]
This can be simplified to:
\[ m = -\frac{a \tan \theta}{b \sec \theta} \]

### Step 4: Write the equation of the line through the center
The center of the hyperbola is at the origin (0,0). The equation of the line perpendicular to the normal can be written as:
\[ y = -\frac{1}{m} x \]
Substituting the value of \(m\):
\[ y = \frac{b}{a} \sin \theta \cdot x \]
Let’s denote this equation as (2).

### Step 5: Substitute the values into equation (1)
From equation (1), substituting \(\sec \theta = \frac{1}{\cos \theta}\) and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):
\[ \frac{ax_0 \cos \theta}{1} + \frac{by_0 \cos \theta}{\sin \theta} = a^2 + b^2 \]

### Step 6: Find the value of \(\sin \theta\)
From equation (2), we have:
\[ \sin \theta = \frac{by}{ax} \]
Substituting this back into the equation gives:
\[ \frac{ax_0 \cos \theta}{1} + \frac{b \left(\frac{by}{ax}\right) \cos \theta}{\frac{by}{ax}} = a^2 + b^2 \]

### Step 7: Simplify the equation
After simplification, we can express the relationship between \(x\) and \(y\):
\[ \frac{a^2y^2}{b^2} - x^2 = a^2 + b^2 \]

### Step 8: Rearranging the equation
Rearranging gives us the final locus equation:
\[ a^2y^2 - b^2x^2 = (a^2 + b^2) b^2 \]

### Final Result
Thus, the locus of the foot of the perpendicular from the center upon any normal to the hyperbola is:
\[ a^2y^2 - b^2x^2 = (a^2 + b^2) b^2 \]