The equation of curve passing through (1,1) in which the subtangent is always bisected at the origin is.

Options

A. `y^(2) = x`
B. `2x^(2) - y ^(2) = 1`
C. `x^(2) + y ^(2) = 2`
D. `x+y = 2`

Answer

To find the equation of the curve that passes through the point (1,1) and has the property that its subtangent is always bisected at the origin, we can follow these steps:

### Step 1: Understand the concept of subtangent
The subtangent at a point on the curve is the horizontal distance from the point where the tangent touches the curve to the y-axis. If the curve is represented by \( y = f(x) \), then at a point \( (x_1, y_1) \), the slope of the tangent line is given by \( f'(x_1) \).

### Step 2: Write the expression for the subtangent
The length of the subtangent \( PM \) can be expressed as:
\[ PM = \frac{y_1}{f'(x_1)} \]
where \( y_1 = f(x_1) \).

### Step 3: Condition of the problem
According to the problem, the subtangent is bisected at the origin. This means that the distance from the point \( P(x_1, y_1) \) to the origin is equal to half the length of the subtangent:
\[ \frac{PM}{2} = x_1 \]
Substituting the expression for \( PM \):
\[ \frac{1}{2} \cdot \frac{y_1}{f'(x_1)} = x_1 \]
This simplifies to:
\[ y_1 = 2x_1 f'(x_1) \]

### Step 4: Replace \( y_1 \) with \( f(x_1) \)
Let \( y = f(x) \). Then we can rewrite the equation as:
\[ f(x) = 2x f'(x) \]

### Step 5: Solve the differential equation
This is a first-order differential equation. We can rearrange it:
\[ \frac{f'(x)}{f(x)} = \frac{1}{2x} \]
Integrating both sides:
\[ \int \frac{f'(x)}{f(x)} \, dx = \int \frac{1}{2x} \, dx \]
This gives:
\[ \ln |f(x)| = \frac{1}{2} \ln |x| + C \]
Exponentiating both sides:
\[ f(x) = k \sqrt{x} \]
where \( k = e^C \).

### Step 6: Use the point (1,1) to find \( k \)
Since the curve passes through the point (1,1):
\[ f(1) = k \sqrt{1} = k = 1 \]
Thus, we have:
\[ f(x) = \sqrt{x} \]

### Step 7: Find the equation of the curve
To express this in standard form:
\[ y^2 = x \]

### Conclusion
The equation of the curve is:
\[ y^2 = x \]