The curve that passes through the point `(2, 3)` and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact, is given by

Options

A. `y=(6)/(x)`
B. `x^(2)+y^(2)=13`
C. `((x)/(2))^(2)+((y)/(3))^(2)=2`
D. `2y-3x=0`

Explanation

To find the curve that passes through the point (2, 3) and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact, we can follow these steps:

### Step 1: Set up the equation of the tangent line
Let \( P(x, y) \) be a point on the curve. The equation of the tangent line at this point can be expressed as:
\[ y - y_1 = \frac{dy}{dx}(x - x_1) \]
where \( (x_1, y_1) \) is the point of tangency.

### Step 2: Find the intercepts of the tangent line
The tangent line intersects the x-axis when \( y = 0 \):
\[ 0 - y_1 = \frac{dy}{dx}(x - x_1) \implies x = x_1 - \frac{y_1}{\frac{dy}{dx}} \]
This gives us the x-intercept \( A \left( x_1 - \frac{y_1}{\frac{dy}{dx}}, 0 \right) \).

The tangent line intersects the y-axis when \( x = 0 \):
\[ y - y_1 = \frac{dy}{dx}(0 - x_1) \implies y = y_1 - x_1 \frac{dy}{dx} \]
This gives us the y-intercept \( B \left( 0, y_1 - x_1 \frac{dy}{dx} \right) \).

### Step 3: Use the midpoint property
Since the point \( P(x_1, y_1) \) bisects the segment \( AB \), we can write:
\[ x_1 = \frac{x_1 - \frac{y_1}{\frac{dy}{dx}} + 0}{2} \quad \text{and} \quad y_1 = \frac{0 + y_1 - x_1 \frac{dy}{dx}}{2} \]

From the x-coordinate:
\[ 2x_1 = x_1 - \frac{y_1}{\frac{dy}{dx}} \implies x_1 = -\frac{y_1}{\frac{dy}{dx}} \implies \frac{dy}{dx} = -\frac{y_1}{x_1} \]

From the y-coordinate:
\[ 2y_1 = y_1 - x_1 \frac{dy}{dx} \implies y_1 = -x_1 \frac{dy}{dx} \implies \frac{dy}{dx} = -\frac{y_1}{x_1} \]

### Step 4: Set up the differential equation
From the above, we have:
\[ \frac{dy}{dx} = -\frac{y}{x} \]

### Step 5: Solve the differential equation
This is a separable differential equation. We can separate variables and integrate:
\[ \frac{dy}{y} = -\frac{dx}{x} \]
Integrating both sides gives:
\[ \ln |y| = -\ln |x| + C \]
Exponentiating both sides results in:
\[ |y| = \frac{C}{|x|} \]
This can be rewritten as:
\[ xy = C \]

### Step 6: Determine the constant using the given point
We know the curve passes through the point (2, 3):
\[ 2 \cdot 3 = C \implies C = 6 \]
Thus, the equation of the curve is:
\[ xy = 6 \]

### Step 7: Express in terms of \( y \)
We can express this as:
\[ y = \frac{6}{x} \]

### Final Answer
The equation of the curve is:
\[ y = \frac{6}{x} \]