Let `f(x,y)` be a curve in the x-y plane having the property that distance from the origin of any tangent to the curve is equal to distance of point of contact from the y-axis. It `f(1,2)=0,` then all such possible curves are

Options

A. `x^(2)+y^(2)=5x`
B. `x^(2)-y^(2)=5x`
C. `x^(2)y^(2)=5x`
D. All of these

Answer

To solve the problem, we need to find the curves \( f(x, y) \) that satisfy the given conditions. Here’s a step-by-step solution:

### Step 1: Understand the problem
We are given a curve in the x-y plane such that the distance from the origin to any tangent line at a point on the curve is equal to the distance from the point of contact to the y-axis. We also know that \( f(1, 2) = 0 \).

### Step 2: Set up the tangent line
Let the point of contact on the curve be \( (x, y) \). The slope of the tangent line at this point can be represented as \( \frac{dy}{dx} \). The equation of the tangent line at the point \( (x, y) \) can be expressed as:
\[ y - y_1 = m(x - x_1) \]
where \( m = \frac{dy}{dx} \) and \( (x_1, y_1) = (x, y) \).

Rearranging gives:
\[ y - y = \frac{dy}{dx}(x - x) \]

### Step 3: Calculate the distance from the origin to the tangent line
The distance \( d \) from the origin \( (0, 0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our tangent line, we can express it in the form \( y - y = \frac{dy}{dx}(x - x) \) which can be rearranged to:
\[ \frac{dy}{dx}x - y + y = 0 \]
Thus, \( A = \frac{dy}{dx} \), \( B = -1 \), and \( C = y \).

The distance from the origin to this line is:
\[ d = \frac{|\frac{dy}{dx}(0) - 1(0) + y|}{\sqrt{(\frac{dy}{dx})^2 + (-1)^2}} = \frac{|y|}{\sqrt{(\frac{dy}{dx})^2 + 1}} \]

### Step 4: Calculate the distance from the point of contact to the y-axis
The distance from the point of contact \( (x, y) \) to the y-axis is simply the x-coordinate, which is \( |x| \).

### Step 5: Set the distances equal
According to the problem, these two distances are equal:
\[ \frac{|y|}{\sqrt{(\frac{dy}{dx})^2 + 1}} = |x| \]

### Step 6: Square both sides
Squaring both sides gives:
\[ \frac{y^2}{(\frac{dy}{dx})^2 + 1} = x^2 \]
Rearranging leads to:
\[ y^2 = x^2((\frac{dy}{dx})^2 + 1) \]

### Step 7: Rearranging the equation
This can be rearranged to:
\[ y^2 = x^2 \frac{dy^2}{dx^2} + x^2 \]

### Step 8: Form the differential equation
Rearranging gives us:
\[ x^2 \frac{dy^2}{dx^2} - y^2 + x^2 = 0 \]
This can be simplified to:
\[ \frac{dy}{dx} + \frac{x^2}{2xy} - \frac{y^2}{2xy} = 0 \]

### Step 9: Solve the differential equation
This is a homogeneous differential equation. We can use the substitution \( y = vx \) where \( v = \frac{y}{x} \). Then, we can express \( \frac{dy}{dx} \) in terms of \( v \) and \( x \).

### Step 10: Integrate and find the constant
After integrating, we will find the general solution and then use the condition \( f(1, 2) = 0 \) to find the constant.

### Final Step: Write the final equation
After substituting back, we find that the required curve is:
\[ x^2 + y^2 = 5x \]