The equation of a curve passing through (1,0) for which the product of the abscissa of a point `P` and the intercept made by a normal at `P` on the x-axis equal twice the square of the radius vector of the point `P` is (a) `( b ) (c) (d) x^(( e )2( f ))( g )+( h ) y^(( i )2( j ))( k )=( l ) x^(( m )4( n ))( o ) (p)` (q) (b) `( r ) (s) (t) x^(( u )2( v ))( w )+( x ) y^(( y )2( z ))( a a )=2( b b ) x^(( c c )4( d d ))( e e ) (ff)` (gg) (c) `( d ) (e) (f) x^(( g )2( h ))( i )+( j ) y^(( k )2( l ))( m )=4( n ) x^(( o )4( p ))( q ) (r)` (s) (d) None of these

Options

A. `x^(2)+y^(2)=x^(4)`
B. `x^(2)+y^(2)=2x^(4)`
C. `x^(2)+y^(2)=4x^(4)`
D. None of these

Explanation

To solve the problem, we need to derive the equation of a curve based on the conditions provided. Let's break it down step by step.

### Step 1: Define the Variables
Let the point \( P \) on the curve be represented by coordinates \( (x, y) \). The abscissa of point \( P \) is \( x \) and the radius vector \( r \) is given by:
\[ r = \sqrt{x^2 + y^2} \]

### Step 2: Find the Normal at Point \( P \)
The slope of the tangent at point \( P \) is given by \( \frac{dy}{dx} \). Therefore, the slope of the normal is:
\[ -\frac{dx}{dy} \]
Using the point-slope form of the equation of a line, the equation of the normal at point \( P \) is:
\[ y - y = -\frac{dx}{dy}(x - x) \]
Rearranging gives:
\[ y = -\frac{dx}{dy}(x - x) + y \]

### Step 3: Find the Intercept on the X-axis
To find the intercept on the x-axis, we set \( y = 0 \):
\[ 0 - y = -\frac{dx}{dy}(x - x) \]
This simplifies to:
\[ y \frac{dy}{dx} = -x \]
From this, we can solve for the intercept \( I \):
\[ I = y \frac{dy}{dx} + x \]

### Step 4: Set Up the Equation Based on Given Condition
According to the problem, the product of the abscissa \( x \) and the intercept \( I \) equals twice the square of the radius vector:
\[ x \cdot I = 2r^2 \]
Substituting for \( I \) and \( r \):
\[ x \left( y \frac{dy}{dx} + x \right) = 2(x^2 + y^2) \]

### Step 5: Rearranging the Equation
Expanding and rearranging gives:
\[ xy \frac{dy}{dx} + x^2 = 2x^2 + 2y^2 \]
This simplifies to:
\[ xy \frac{dy}{dx} = 2x^2 + 2y^2 - x^2 \]
\[ xy \frac{dy}{dx} = x^2 + 2y^2 \]

### Step 6: Separate Variables
Rearranging gives:
\[ \frac{dy}{dx} = \frac{x^2 + 2y^2}{xy} \]

### Step 7: Homogeneous Equation
To solve this, we can use the substitution \( y = vx \) (where \( v = \frac{y}{x} \)):
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]
Substituting this into the equation gives:
\[ v + x \frac{dv}{dx} = \frac{x^2 + 2(vx)^2}{x(vx)} = \frac{x^2 + 2v^2x^2}{vx^2} \]
This simplifies to:
\[ v + x \frac{dv}{dx} = \frac{1 + 2v^2}{v} \]

### Step 8: Solve the Differential Equation
Rearranging gives:
\[ x \frac{dv}{dx} = \frac{1 + 2v^2 - v^2}{v} = \frac{1 + v^2}{v} \]
This can be integrated to find \( v \) in terms of \( x \).

### Step 9: Integrate
Integrate both sides:
\[ \int \frac{v}{1 + v^2} dv = \int \frac{1}{x} dx \]
This leads to logarithmic forms that can be solved to find the relationship between \( x \) and \( y \).

### Step 10: Final Equation
After integrating and manipulating, we arrive at the final equation, which is:
\[ x^2 + y^2 = x^4 \]

### Conclusion
The equation of the curve passing through the point \( (1, 0) \) is:
\[ x^2 + y^2 = x^4 \]