The locus of the foot of the perpendicular from the center of the hyperbola `x y=1` on a variable tangent is (a) `(x^(2)+y^(2))^(2)=4xy` (b) `(x^2-y^2)=1/9` (c) `(x^2-y^2)=7/(144)` (d) `(x^2-y^2)=1/(16)`

Options

A. `(x^(2)-y^(2))^(2)=4xy`
B. `(x^(2)+y^(2))^(2)=2xy`
C. `(x^(2)+y^(2))=4xy`
D. `(x^(2)+y^(2))^(2)=4xy`

Explanation

To find the locus of the foot of the perpendicular from the center of the hyperbola \( xy = 1 \) on a variable tangent, we'll follow these steps:

### Step 1: Identify the center of the hyperbola
The hyperbola given is \( xy = 1 \). The center of this hyperbola is at the origin, \( (0, 0) \).

**Hint:** The center of a hyperbola in the form \( xy = c \) is always at the origin.

### Step 2: Determine the slope of the perpendicular from the center
Let the foot of the perpendicular from the center to the tangent be a point \( P(h, k) \). The slope of the line from the center \( O(0, 0) \) to the point \( P(h, k) \) is given by:

\[ \text{slope of OP} = \frac{k}{h} \]

Since the tangent at point \( P \) is perpendicular to this line, the slope of the tangent line will be:

\[ \text{slope of tangent} = -\frac{h}{k} \]

**Hint:** Remember that perpendicular slopes multiply to -1.

### Step 3: Write the equation of the tangent line
Using the point-slope form of the equation of a line, the equation of the tangent at point \( P(h, k) \) can be expressed as:

\[ y - k = -\frac{h}{k}(x - h) \]

Rearranging this gives:

\[ k(y - k) = -h(x - h) \]

This simplifies to:

\[ hx + ky = h^2 + k^2 \]

**Hint:** Use the standard form of the line equation to rearrange terms.

### Step 4: Substitute the hyperbola equation
From the hyperbola \( xy = 1 \), we can express \( y \) in terms of \( x \):

\[ y = \frac{1}{x} \]

Substituting this into the tangent equation:

\[ hx + k\left(\frac{1}{x}\right) = h^2 + k^2 \]

Multiplying through by \( x \) to eliminate the fraction gives:

\[ hx^2 + k = (h^2 + k^2)x \]

Rearranging this leads to:

\[ hx^2 - (h^2 + k^2)x + k = 0 \]

**Hint:** This is a quadratic equation in \( x \).

### Step 5: Use the condition for tangency
For the line to be tangent to the hyperbola, the discriminant of this quadratic must be zero:

\[ D = (h^2 + k^2)^2 - 4hk = 0 \]

This simplifies to:

\[ (h^2 + k^2)^2 = 4hk \]

**Hint:** The condition for tangency involves setting the discriminant to zero.

### Step 6: Express in terms of \( h \) and \( k \)
Rearranging gives:

\[ h^2 + k^2 = 2\sqrt{hk} \]

Squaring both sides leads to:

\[ (h^2 + k^2)^2 = 4hk \]

### Step 7: Identify the locus
The equation \( (h^2 + k^2)^2 = 4hk \) represents the locus of the foot of the perpendicular from the center of the hyperbola on the tangent.

This can be rewritten in terms of \( x \) and \( y \):

\[ (x^2 + y^2)^2 = 4xy \]

### Final Answer
Thus, the locus of the foot of the perpendicular from the center of the hyperbola \( xy = 1 \) on a variable tangent is:

\[ \boxed{(x^2 + y^2)^2 = 4xy} \]