Let a curve `y=f(x)` pass through (1,1), at any point p on the curve tangent and normal are drawn to intersect the X-axis at Q and R respectively. If QR = 2 , find the equation of all such possible curves.

Answer

To solve the problem, we need to find the equation of curves \( y = f(x) \) that pass through the point (1, 1) and satisfy the condition that the distance \( QR = 2 \), where \( Q \) and \( R \) are the points where the tangent and normal at any point \( P(x, y) \) on the curve intersect the x-axis.

### Step-by-Step Solution:

1. **Define the slope of the tangent and normal:**
Let \( y = f(x) \). At any point \( P(x, y) \), the slope of the tangent is given by \( \frac{dy}{dx} \) and the slope of the normal is \( -\frac{dx}{dy} \).

2. **Equation of the tangent:**
The equation of the tangent line at point \( P(x, y) \) can be expressed as:
\[ y - y_1 = \frac{dy}{dx}(x - x_1) \]
Replacing \( y_1 \) with \( y \) and \( x_1 \) with \( x \):
\[ y - y = \frac{dy}{dx}(x - x) \implies y = \frac{dy}{dx}(x - x) + y \]

3. **Finding the intersection point \( Q \) on the x-axis:**
Set \( y = 0 \) to find where the tangent intersects the x-axis:
\[ 0 - y = \frac{dy}{dx}(X - x) \implies X = x - \frac{y}{\frac{dy}{dx}} \]
Thus, the coordinates of point \( Q \) are:
\[ Q\left(x - \frac{y}{\frac{dy}{dx}}, 0\right) \]

4. **Equation of the normal:**
The equation of the normal at point \( P(x, y) \) is:
\[ y - y = -\frac{dx}{dy}(X - x) \implies y = -\frac{dx}{dy}(X - x) + y \]
Setting \( y = 0 \) to find the x-intercept \( R \):
\[ 0 - y = -\frac{dx}{dy}(X - x) \implies X = x + \frac{y}{\frac{dx}{dy}} \]
Thus, the coordinates of point \( R \) are:
\[ R\left(x + \frac{y}{\frac{dx}{dy}}, 0\right) \]

5. **Finding the distance \( QR \):**
The distance \( QR \) is given by:
\[ QR = X_R - X_Q = \left(x + \frac{y}{\frac{dx}{dy}}\right) - \left(x - \frac{y}{\frac{dy}{dx}}\right) = \frac{y}{\frac{dx}{dy}} + \frac{y}{\frac{dy}{dx}} = y \left(\frac{1}{\frac{dx}{dy}} + \frac{1}{\frac{dy}{dx}}\right) \]
Since \( QR = 2 \):
\[ y \left(\frac{1}{\frac{dx}{dy}} + \frac{1}{\frac{dy}{dx}}\right) = 2 \]

6. **Using the relationship between derivatives:**
Note that \( \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} \). Therefore:
\[ y \left(\frac{dy}{dx} + \frac{1}{\frac{dy}{dx}}\right) = 2 \]
Let \( m = \frac{dy}{dx} \):
\[ y \left(m + \frac{1}{m}\right) = 2 \]
This simplifies to:
\[ y \left(\frac{m^2 + 1}{m}\right) = 2 \]
Rearranging gives:
\[ y(m^2 + 1) = 2m \]

7. **Separating variables:**
Rearranging gives:
\[ y = \frac{2m}{m^2 + 1} \]
We can express \( m \) in terms of \( y \):
\[ m^2 + 1 = \frac{2m}{y} \implies m^2 - \frac{2m}{y} + 1 = 0 \]
Solving this quadratic equation for \( m \):
\[ m = \frac{2/y \pm \sqrt{(2/y)^2 - 4}}{2} \]

8. **Integrating to find the curve:**
Integrate \( \frac{dy}{dx} = m \) to find the equation of the curve. The curves that satisfy the initial condition \( f(1) = 1 \) will be:
\[ y^2 = 2x - 1 \quad \text{and} \quad y^2 = -2x + 3 \]

### Final Result:
The equations of the curves that satisfy the given conditions are:
1. \( y^2 = 2x - 1 \)
2. \( y^2 = -2x + 3 \)