The perpendicular from the origin to the tangent at any point on a curve is equal to the abscissa of the point of contact. Also curve passes through the point (1,1). Then the length of intercept of the curve on the x-axis is__________

Explanation

To solve the problem step by step, we will derive the equation of the curve based on the given conditions and then find the length of the intercept on the x-axis.

### Step 1: Understand the Given Condition
We know that the perpendicular distance from the origin to the tangent at any point \( P(h, k) \) on the curve is equal to the abscissa \( h \) of the point of contact.

### Step 2: Equation of the Tangent
The slope of the tangent at point \( P(h, k) \) is given by \( \frac{dy}{dx} \). The equation of the tangent line can be expressed as:
\[ y - k = \frac{dy}{dx}(x - h) \]
Rearranging gives us:
\[ \frac{dy}{dx}(x - h) - (y - k) = 0 \]

### Step 3: Perpendicular Distance from the Origin
The perpendicular distance \( d \) from the origin (0, 0) to the tangent line can be calculated using the formula:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our tangent line, \( A = \frac{dy}{dx} \), \( B = -1 \), and \( C = k - \frac{dy}{dx}h \). Thus,
\[ d = \frac{|\frac{dy}{dx}(0) - 1(0) + (k - \frac{dy}{dx}h)|}{\sqrt{(\frac{dy}{dx})^2 + (-1)^2}} = \frac{|k - \frac{dy}{dx}h|}{\sqrt{(\frac{dy}{dx})^2 + 1}} \]
According to the problem, this distance \( d \) is equal to \( h \):
\[ \frac{|k - \frac{dy}{dx}h|}{\sqrt{(\frac{dy}{dx})^2 + 1}} = h \]

### Step 4: Squaring Both Sides
Squaring both sides to eliminate the absolute value gives us:
\[ (k - \frac{dy}{dx}h)^2 = h^2((\frac{dy}{dx})^2 + 1) \]

### Step 5: Expand and Rearrange
Expanding both sides:
\[ k^2 - 2k\frac{dy}{dx}h + h^2\left(\frac{dy}{dx}\right)^2 = h^2\left(\frac{dy}{dx}\right)^2 + h^2 \]
Cancelling \( h^2\left(\frac{dy}{dx}\right)^2 \) from both sides gives:
\[ k^2 - 2k\frac{dy}{dx}h = h^2 \]

### Step 6: Solve for \(\frac{dy}{dx}\)
Rearranging gives:
\[ 2k\frac{dy}{dx}h = k^2 - h^2 \]
Thus,
\[ \frac{dy}{dx} = \frac{k^2 - h^2}{2kh} \]

### Step 7: Generalizing to \( (x, y) \)
Substituting \( h \) with \( x \) and \( k \) with \( y \):
\[ \frac{dy}{dx} = \frac{y^2 - x^2}{2xy} \]

### Step 8: Solve the Differential Equation
This is a homogeneous differential equation. We can use the substitution \( y = vx \) (where \( v = \frac{y}{x} \)):
\[ \frac{dy}{dx} = v + x\frac{dv}{dx} \]
Substituting this into the differential equation gives:
\[ v + x\frac{dv}{dx} = \frac{v^2x^2 - x^2}{2vx} = \frac{x^2(v^2 - 1)}{2vx} \]
Simplifying leads to:
\[ x\frac{dv}{dx} = \frac{-v^2 + 1}{2v} \]

### Step 9: Separate Variables
This can be separated as:
\[ \frac{2v}{1 - v^2} dv = -\frac{1}{x} dx \]

### Step 10: Integrate Both Sides
Integrating both sides:
\[ \int \frac{2v}{1 - v^2} dv = -\int \frac{1}{x} dx \]
The left side integrates to \( -\ln|1 - v^2| \) and the right side to \( -\ln|x| + C \).

### Step 11: Substitute Back
Substituting back \( v = \frac{y}{x} \):
\[ -\ln|1 - \frac{y^2}{x^2}| = -\ln|x| + C \]
Exponentiating gives:
\[ \frac{1}{1 - \frac{y^2}{x^2}} = \frac{k}{x} \]
where \( k = e^C \).

### Step 12: Find the X-Intercept
To find the x-intercept, set \( y = 0 \):
\[ \frac{x}{2} = 1 \implies x = 2 \]

### Final Answer
The length of the intercept of the curve on the x-axis is **2**.