A curve is such that the x - intercept of the tangent drawn to it the point P(x, y) is reciprocal of the abscissa of P. Then, the equation of the curveis (where, c is the constant of integration and `x gt 1`)

Options

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

Answer

To solve the problem, we need to find the equation of a curve given that the x-intercept of the tangent drawn to the curve at a point \( P(x, y) \) is the reciprocal of the abscissa of \( P \).

### Step-by-step Solution:

1. **Understanding the Problem**:
We know that the x-intercept of the tangent line at point \( P(x, y) \) is given by \( x_1 = \frac{1}{x} \).

2. **Finding the Slope of the Tangent**:
The slope of the tangent at point \( P(x, y) \) can be expressed as \( \frac{dy}{dx} \).

3. **Using the Point-Slope Form**:
The equation of the tangent line at point \( P(x, y) \) can be written using the point-slope form:
\[ y - y_1 = m(x - x_1) \]
where \( m = \frac{dy}{dx} \) and \( (x_1, y_1) = (x, y) \).

4. **Finding the x-intercept**:
The x-intercept occurs when \( y = 0 \). Setting \( y = 0 \) in the tangent equation gives:
\[ 0 - y = \frac{dy}{dx}(x - x) \]
Rearranging gives:
\[ x_1 = x - \frac{y}{\frac{dy}{dx}} \]
Since we know \( x_1 = \frac{1}{x} \), we can set the two expressions for \( x_1 \) equal:
\[ \frac{1}{x} = x - \frac{y}{\frac{dy}{dx}} \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{y}{\frac{dy}{dx}} = x - \frac{1}{x} \]
Thus, we can express \( \frac{dy}{dx} \) as:
\[ \frac{dy}{dx} = \frac{y}{x - \frac{1}{x}} \]

6. **Simplifying the Expression**:
This can be rewritten as:
\[ \frac{dy}{dx} = \frac{y}{\frac{x^2 - 1}{x}} = \frac{xy}{x^2 - 1} \]

7. **Separating Variables**:
We can separate variables to integrate:
\[ \frac{1}{y} dy = \frac{x}{x^2 - 1} dx \]

8. **Integrating Both Sides**:
Integrating both sides:
\[ \int \frac{1}{y} dy = \int \frac{x}{x^2 - 1} dx \]
The left side integrates to \( \ln |y| \). For the right side, we can use substitution:
Let \( t = x^2 - 1 \), then \( dt = 2x dx \) or \( dx = \frac{dt}{2x} \):
\[ \int \frac{x}{t} \cdot \frac{dt}{2x} = \frac{1}{2} \int \frac{1}{t} dt = \frac{1}{2} \ln |t| + C \]

9. **Combining the Results**:
Thus, we have:
\[ \ln |y| = \frac{1}{2} \ln |x^2 - 1| + C \]
Exponentiating both sides gives:
\[ |y| = e^C (x^2 - 1)^{1/2} \]
Let \( k = e^C \), then:
\[ y = k \sqrt{x^2 - 1} \]

10. **Final Equation**:
Therefore, the equation of the curve is:
\[ y = c \sqrt{x^2 - 1} \]
where \( c \) is a constant and \( x > 1 \).