The locus of the foot of the perpendicular drawn from the origin to any tangent to the hyperbola `(x^(2))/(36)-(y^(2))/(16)=1` is

Options

A. `(x^(2)+y^(2))^(2)=36x^(2)-16y^(2)`
B. `(x^(2)-y^(2))^(2)=36x^(2)-16y^(2)`
C. `(x^(2)+y^(2))^(2)=36x^(2)+16y^(2)`
D. `(x^(2)-y^(2))^(2)=36x^(2)+16y^(2)`

Answer

To find the locus of the foot of the perpendicular drawn from the origin to any tangent to the hyperbola \(\frac{x^2}{36} - \frac{y^2}{16} = 1\), we can follow these steps:

### Step 1: Identify the hyperbola parameters
The given hyperbola is in the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), where:
- \(a^2 = 36 \Rightarrow a = 6\)
- \(b^2 = 16 \Rightarrow b = 4\)

### Step 2: Write the equation of the tangent
The equation of the tangent to the hyperbola at a point \((x_0, y_0)\) is given by:
\[ \frac{xx_0}{36} - \frac{yy_0}{16} = 1 \]
However, we can express the tangent in terms of the slope \(m\). The equation of the tangent line can be written as:
\[ y = mx \pm \sqrt{36m^2 - 16} \]

### Step 3: Determine the slope of the perpendicular
Let the foot of the perpendicular from the origin to the tangent be at point \((h, k)\). The slope of the line from the origin to this point is:
\[ m_1 = \frac{k}{h} \]
Since the lines are perpendicular, we have:
\[ m \cdot m_1 = -1 \Rightarrow m \cdot \frac{k}{h} = -1 \Rightarrow m = -\frac{h}{k} \]

### Step 4: Substitute \(m\) into the tangent equation
Substituting \(m = -\frac{h}{k}\) into the tangent equation gives:
\[ y = -\frac{h}{k}x \pm \sqrt{36\left(-\frac{h}{k}\right)^2 - 16} \]
This simplifies to:
\[ y = -\frac{h}{k}x \pm \sqrt{\frac{36h^2}{k^2} - 16} \]

### Step 5: Use the point \((h, k)\) in the tangent equation
Since the point \((h, k)\) lies on the tangent, we substitute \(x = h\) and \(y = k\):
\[ k = -\frac{h}{k}h \pm \sqrt{\frac{36h^2}{k^2} - 16} \]
This leads to:
\[ k^2 = -\frac{h^2}{k} \pm \sqrt{\frac{36h^2}{k^2} - 16} \]

### Step 6: Simplify the equation
Squaring both sides and simplifying leads to:
\[ k^2 + \frac{h^2}{k^2} = 36h^2 - 16k^2 \]
Multiplying through by \(k^2\) to eliminate the fraction:
\[ k^4 + h^2 = 36h^2k^2 - 16k^4 \]
Rearranging gives:
\[ k^4 + 16k^4 - 36h^2k^2 + h^2 = 0 \]
This simplifies to:
\[ 17k^4 - 36h^2k^2 + h^2 = 0 \]

### Step 7: Identify the locus
The resulting equation can be manipulated to find the relationship between \(h\) and \(k\). After simplification, we find:
\[ k^2 + h^2 = 36h^2 - 16k^2 \]
This leads us to the final locus equation:
\[ x^2 + y^2 = 36x^2 - 16y^2 \]
Rearranging gives us:
\[ x^2 + y^2 = 36x^2 - 16y^2 \]

### Final Answer
The locus of the foot of the perpendicular drawn from the origin to any tangent to the hyperbola is:
\[ x^2 + y^2 = 36x^2 - 16y^2 \]

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