The feet of the normals to `(x^(2))/(a^(2)) -(y^(2))/(b^(2)) =1` from `(h,k)` lie on

Options

A. (a)`a^(2)y(x-h) + b^(2)x(y-k) = 0`
B. (b)`b^(2)x(x-h) + a^(2)y(y-k) =0`
C. (c)`(a^(2) + b^(2)) xy -a^(2)hy -b^(2)xk = 0`
D. (d)None of these

Answer

To find the feet of the normals to the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) from the point \((h, k)\), we can follow these steps:

### Step-by-Step Solution

1. **Identify the equation of the hyperbola**:
The given hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

2. **Write the equation of the normal**:
The equation of the normal to the hyperbola at the point \((\alpha, \beta)\) can be expressed as:
\[ \frac{a^2 x}{\alpha} + \frac{b^2 y}{\beta} = a^2 + b^2 \]

3. **Substitute the point \((h, k)\)**:
Since the normal passes through the point \((h, k)\), we substitute \(x = h\) and \(y = k\) into the normal equation:
\[ \frac{a^2 h}{\alpha} + \frac{b^2 k}{\beta} = a^2 + b^2 \]

4. **Multiply through by \(\alpha \beta\)** to eliminate the denominators:
\[ a^2 h \beta + b^2 k \alpha = (a^2 + b^2) \alpha \beta \]

5. **Rearranging the equation**:
Rearranging gives us:
\[ a^2 h \beta + b^2 k \alpha - (a^2 + b^2) \alpha \beta = 0 \]

6. **Substitute \(\alpha\) and \(\beta\) with \(x\) and \(y\)**:
Let \(\alpha = x\) and \(\beta = y\):
\[ a^2 h y + b^2 k x - (a^2 + b^2) xy = 0 \]

7. **Rearranging the equation**:
This can be rearranged to:
\[ a^2 h y + b^2 k x = (a^2 + b^2) xy \]

8. **Final form**:
The final equation represents the locus of the feet of the normals from the point \((h, k)\):
\[ b^2 x (y - k) + a^2 y (x - h) = 0 \]

### Conclusion
The feet of the normals to the hyperbola from the point \((h, k)\) lie on the curve defined by:
\[ b^2 x (y - k) + a^2 y (x - h) = 0 \]