The curve for which normal at any point (x, y) and the line joining origin to that point form an isosceles triangle with x-axis is

Options

A. circle
B. parabola
C. ellipse
D. rectangular hyperbola

Answer

To find the curve for which the normal at any point \((x, y)\) and the line joining the origin to that point form an isosceles triangle with the x-axis, we can follow these steps:

### Step 1: Understand the Geometry
The triangle formed by the normal line at point \((x, y)\), the line from the origin \((0, 0)\) to \((x, y)\), and the x-axis needs to be isosceles. This means that the lengths of the two sides of the triangle formed by the normal and the line from the origin must be equal.

### Step 2: Find the Slope of the Tangent and Normal
Let \(y = f(x)\) be the equation of the curve. The slope of the tangent line at point \((x, y)\) is given by \(f'(x)\). Therefore, the slope of the normal line, which is perpendicular to the tangent, is \(-\frac{1}{f'(x)}\).

### Step 3: Equation of the Normal Line
The equation of the normal line at point \((x, y)\) can be expressed as:
\[ y - f(x) = -\frac{1}{f'(x)}(x - x_0) \]
This can be rearranged to:
\[ y = -\frac{1}{f'(x)}(x - x_0) + f(x) \]

### Step 4: Find the Intersection with the X-axis
To find where the normal intersects the x-axis, set \(y = 0\):
\[ 0 = -\frac{1}{f'(x)}(x - x_0) + f(x) \]
Solving for \(x\) gives:
\[ \frac{1}{f'(x)}(x - x_0) = f(x) \]
\[ x - x_0 = f(x) f'(x) \]
\[ x = x_0 + f(x) f'(x) \]

### Step 5: Lengths of the Sides of the Triangle
The lengths of the sides of the triangle formed by the normal and the line from the origin to \((x, y)\) must be equal. The length from the origin to the point \((x, y)\) is:
\[ L_1 = \sqrt{x^2 + f(x)^2} \]
The length from the origin to the intersection point on the x-axis is:
\[ L_2 = |x_0| = |x_0 + f(x) f'(x)| \]

### Step 6: Set Up the Isosceles Condition
For the triangle to be isosceles, we need:
\[ \sqrt{x^2 + f(x)^2} = |x + f(x) f'(x)| \]
Squaring both sides to eliminate the square root gives:
\[ x^2 + f(x)^2 = (x + f(x) f'(x))^2 \]

### Step 7: Expand and Simplify
Expanding the right-hand side:
\[ x^2 + f(x)^2 = x^2 + 2x f(x) f'(x) + f(x)^2 f'(x)^2 \]
Subtracting \(x^2 + f(x)^2\) from both sides:
\[ 0 = 2x f(x) f'(x) + f(x)^2 f'(x)^2 \]

### Step 8: Factor Out Common Terms
Factoring out \(f(x)\):
\[ f(x)(2x f'(x) + f(x) f'(x)^2) = 0 \]

### Step 9: Solve the Equation
This gives us two cases:
1. \(f(x) = 0\) (which is trivial and not a curve)
2. \(2x f'(x) + f(x) f'(x)^2 = 0\)

### Step 10: Solve the Differential Equation
Rearranging gives:
\[ f'(x)(2x + f(x) f'(x)) = 0 \]
This leads to a separable differential equation. Solving this will yield the required curve.

### Final Solution
The solution to the differential equation will provide the equation of the curve.