Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola `16y^2 -9 x^2 = 1` is

Options

A. `x^(2)+y^(2)=9`
B. `x^(2)+y^(2)=1//9`
C. `x^(2)+y^(2)=7//144`
D. `x^(2)+y^(2)=1//16`

Explanation

To find the locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola given by the equation \(16y^2 - 9x^2 = 1\), we will follow these steps:

### Step 1: Rewrite the Hyperbola Equation
The given hyperbola equation is \(16y^2 - 9x^2 = 1\). We can rewrite it in standard form:
\[ \frac{y^2}{\frac{1}{16}} - \frac{x^2}{\frac{1}{9}} = 1 \]
This shows that \(a^2 = \frac{1}{16}\) and \(b^2 = \frac{1}{9}\).

### Step 2: Identify the Foci
The foci of the hyperbola can be found using the formula \(c = \sqrt{a^2 + b^2}\):
\[ c = \sqrt{\frac{1}{16} + \frac{1}{9}} = \sqrt{\frac{9 + 16}{144}} = \sqrt{\frac{25}{144}} = \frac{5}{12} \]
Thus, the foci are located at \((0, \pm \frac{5}{12})\).

### Step 3: Equation of the Tangent
The equation of a tangent to the hyperbola can be expressed as:
\[ \frac{y}{\frac{1}{4}} - \frac{x}{\frac{1}{3}} = t \]
where \(t\) is a parameter. Rearranging gives:
\[ y = \frac{1}{4}t + \frac{1}{3}tx \]

### Step 4: Find the Feet of the Perpendiculars
To find the feet of the perpendiculars from the foci \((0, \frac{5}{12})\) and \((0, -\frac{5}{12})\) to the tangent line, we need to use the formula for the foot of the perpendicular from a point to a line.

The general form of the line is \(Ax + By + C = 0\). For our tangent line, we can express it in this form and find the perpendicular distance from the foci.

### Step 5: Locus of the Feet of the Perpendiculars
The locus of the feet of the perpendiculars from the foci can be derived using the property of hyperbolas. The locus will be a circle given by:
\[ x^2 + y^2 = \frac{a^2}{b^2} = \frac{\frac{1}{16}}{\frac{1}{9}} = \frac{9}{16} \]
Thus, the equation simplifies to:
\[ x^2 + y^2 = \frac{1}{16} \]

### Final Answer
The locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola is:
\[ x^2 + y^2 = \frac{1}{16} \]