The tangent at any point P of a curve meets the axis of x in T. The curve for which OP= PT,O being the origin is

Options

A. circle
B. parabola
C. ellipse
D. None of these

Answer

To solve the problem, we need to find the equation of the curve such that the distance from the origin \( O \) to the point \( P \) on the curve is equal to the distance from the point \( P \) to the point \( T \), where \( T \) is the point where the tangent at \( P \) meets the x-axis.

### Step 1: Define the curve
Let the curve be represented by the function \( y = f(x) \). The point \( P \) on the curve can be represented as \( P(x, f(x)) \).

### Step 2: Find the slope of the tangent
The slope of the tangent at point \( P \) is given by the derivative \( f'(x) \). Therefore, the equation of the tangent line at point \( P \) can be written as:
\[ y - f(x) = f'(x)(x - x_0) \]
where \( (x_0, y_0) = (x, f(x)) \).

### Step 3: Find the x-intercept of the tangent line
To find the x-intercept \( T \) of the tangent line, we set \( y = 0 \):
\[ 0 - f(x) = f'(x)(x - x_0) \]
Solving for \( x \) gives:
\[ f'(x)(x - x_0) = -f(x) \]
\[ x - x_0 = -\frac{f(x)}{f'(x)} \]
\[ x = x_0 - \frac{f(x)}{f'(x)} \]

### Step 4: Calculate the distances
The distance \( OP \) from the origin \( O(0, 0) \) to point \( P(x, f(x)) \) is:
\[ OP = \sqrt{x^2 + (f(x))^2} \]

The distance \( PT \) from point \( P \) to point \( T \) is:
\[ PT = \sqrt{(x - (x_0 - \frac{f(x)}{f'(x)}))^2 + (f(x) - 0)^2} \]
This simplifies to:
\[ PT = \sqrt{\left(\frac{f(x)}{f'(x)}\right)^2 + (f(x))^2} \]

### Step 5: Set the distances equal
According to the problem, we have:
\[ OP = PT \]
Substituting the expressions we obtained:
\[ \sqrt{x^2 + (f(x))^2} = \sqrt{\left(\frac{f(x)}{f'(x)}\right)^2 + (f(x))^2} \]

### Step 6: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ x^2 + (f(x))^2 = \left(\frac{f(x)}{f'(x)}\right)^2 + (f(x))^2 \]
This simplifies to:
\[ x^2 = \left(\frac{f(x)}{f'(x)}\right)^2 \]

### Step 7: Rearranging the equation
Rearranging gives us:
\[ f'(x)^2 = \frac{f(x)^2}{x^2} \]
Taking the square root of both sides:
\[ f'(x) = \frac{f(x)}{x} \]

### Step 8: Solve the differential equation
This is a separable differential equation. We can rewrite it as:
\[ \frac{f'(x)}{f(x)} = \frac{1}{x} \]
Integrating both sides gives:
\[ \ln |f(x)| = \ln |x| + C \]
Exponentiating both sides results in:
\[ f(x) = kx \]
where \( k = e^C \) is a constant.

### Conclusion
The curve for which \( OP = PT \) is a straight line through the origin, expressed as:
\[ y = kx \]
where \( k \) is a constant.