A curve passing through the point(1, 1) is such that the intercept made by a tangent to it on x-axis is three times the x co-ordinate of the point of tangency, then the equation of the curve is :

Options

A. `y=(1)/(x^(2))`
B. `y=sqrt(x)`
C. `y=(1)/(sqrt(x))`
D. none

Answer

To solve the problem, we need to find the equation of a curve given that the intercept made by a tangent to the curve on the x-axis is three times the x-coordinate of the point of tangency. The curve also passes through the point (1, 1).

Let's denote the curve by \( y = f(x) \).

### Step 1: Find the equation of the tangent line
The slope of the tangent line at a point \( (x_0, y_0) \) on the curve is given by \( f'(x_0) \). The equation of the tangent line at the point \( (x_0, f(x_0)) \) can be written as:

\[ y - f(x_0) = f'(x_0)(x - x_0) \]

### Step 2: Find the x-intercept of the tangent line
To find the x-intercept, we set \( y = 0 \):

\[ 0 - f(x_0) = f'(x_0)(x - x_0) \]

Solving for \( x \):

\[ f'(x_0)(x - x_0) = f(x_0) \]
\[ x - x_0 = \frac{f(x_0)}{f'(x_0)} \]
\[ x = x_0 + \frac{f(x_0)}{f'(x_0)} \]

### Step 3: Set up the relationship given in the problem
According to the problem, the x-intercept is three times the x-coordinate of the point of tangency:

\[ x_0 + \frac{f(x_0)}{f'(x_0)} = 3x_0 \]

Rearranging gives:

\[ \frac{f(x_0)}{f'(x_0)} = 3x_0 - x_0 = 2x_0 \]

### Step 4: Separate variables and integrate
This can be rewritten as:

\[ f(x_0) = 2x_0 f'(x_0) \]

Dividing both sides by \( f(x_0) \):

\[ \frac{f'(x_0)}{f(x_0)} = \frac{1}{2x_0} \]

Now, we integrate 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 \( k = e^C \) is a constant.

### Step 5: Use the initial condition
Since the curve passes through the point (1, 1):

\[ f(1) = k \sqrt{1} = k = 1 \]

Thus, the equation of the curve is:

\[ f(x) = \sqrt{x} \]

### Final Answer
The equation of the curve is:

\[ y = \sqrt{x} \]

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