Find the locus of feet of perpendicular from (5,0) to the tangents of `(x^(2))/(16)-y^(2)/(9)=1`

Options

A. `x^(2)+y^(2)=4`
B. `x^(2)+y^(2)=16`
C. `x^(2)+y^(2)=9`
D. `x^(2)+y^(2)=25`

Answer

To find the locus of the foot of the perpendicular from the point (5, 0) to the tangents of the hyperbola given by the equation \(\frac{x^2}{16} - \frac{y^2}{9} = 1\), we will follow these steps:

### Step 1: Identify the hyperbola parameters
The equation of the hyperbola is \(\frac{x^2}{16} - \frac{y^2}{9} = 1\). Here, we can identify:
- \(a^2 = 16\) which gives \(a = 4\),
- \(b^2 = 9\) which gives \(b = 3\).

### Step 2: Parametric representation of the hyperbola
The parametric coordinates of a point on the hyperbola can be expressed as:
\[ P(t) = (a \sec \theta, b \tan \theta) = (4 \sec \theta, 3 \tan \theta) \]
Let this point be \(P\).

### Step 3: Equation of the tangent at point P
The equation of the tangent to the hyperbola at point \(P\) is given by:
\[ \frac{x \sec \theta}{4} - \frac{y \tan \theta}{3} = 1 \]
Multiplying through by 12 (the least common multiple of the denominators) gives:
\[ 3x \sec \theta - 4y \tan \theta = 12 \]

### Step 4: Foot of the perpendicular from (5, 0) to the tangent
Let the foot of the perpendicular from the point (5, 0) to the tangent be denoted by \(A(h, k)\). The coordinates of \(A\) can be found using the formula for the foot of the perpendicular from a point to a line. The line has the equation:
\[ 3x \sec \theta - 4y \tan \theta - 12 = 0 \]
Using the formula for the foot of the perpendicular:
\[ \frac{h - 5}{3 \sec \theta} = \frac{k - 0}{-4 \tan \theta} = -\frac{15 \sec \theta - 12}{9 \sec^2 \theta + 16 \tan^2 \theta} \]

### Step 5: Simplifying the equations
From the equations derived, we can express \(h\) and \(k\) in terms of \(\theta\):
1. From the first equality:
\[ h - 5 = -\frac{9 \sec \theta}{5 \sec \theta + 4} \]
Rearranging gives:
\[ h = -\frac{9 \sec \theta}{5 \sec \theta + 4} + 5 \]

2. From the second equality:
\[ k = -\frac{12 \tan \theta}{5 \sec \theta + 4} \]

### Step 6: Finding the locus
Now, we square both expressions for \(h\) and \(k\) and add them:
\[ h^2 + k^2 = \left(-\frac{9 \sec \theta + 20}{5 \sec \theta + 4}\right)^2 + \left(-\frac{12 \tan \theta}{5 \sec \theta + 4}\right)^2 \]
After simplification, we find that:
\[ h^2 + k^2 = 16 \]

### Conclusion
Thus, the locus of the foot of the perpendicular from the point (5, 0) to the tangents of the hyperbola is given by the equation:
\[ x^2 + y^2 = 16 \]