Tangent to a curve intercepts the y-axis at a point `Pdot` A line perpendicular to this tangent through `P` passes through another point (1,0). The differential equation of the curve is (a) `( b ) (c) y (d)(( e ) dy)/( f )(( g ) dx)( h ) (i)-x (j) (k)(( l ) (m) (n)(( o ) dy)/( p )(( q ) dx)( r ) (s) (t))^(( u )2( v ))( w )=1( x )` (y) (b) `( z ) (aa) (bb)(( c c ) x (dd) d^(( e e )2( f f ))( g g ))/( h h )(( i i ) d (jj) x^(( k k )2( l l ))( m m ))( n n ) (oo)+( p p ) (qq)(( r r ) (ss) (tt)(( u u ) dy)/( v v )(( w w ) dx)( x x ) (yy) (zz))^(( a a a )2( b b b ))( c c c )=0( d d d )` (eee) (c) `( d ) (e) y (f)(( g ) dx)/( h )(( i ) dy)( j ) (k)+x=1( l )` (m) (d) None of these

Options

A. `y(dy)/(dx) -x((dy)/(dx))^(2)=1`
B. `(xd^(2)y)/(dx^(2))+((dy)/(dx))^(2)=0`
C. `y(dy)/(dx)+x=1`
D. None of these

Explanation

To solve the problem, we need to derive the differential equation of the curve based on the given conditions about the tangent and the perpendicular line. Here’s a step-by-step breakdown of the solution:

### Step 1: Define the points and the tangent
Let \( P \) be the point where the tangent to the curve intersects the y-axis. Let the curve be represented by the function \( y = f(x) \). The point \( R \) on the curve where the tangent is drawn can be represented as \( (x, f(x)) \).

### Step 2: Write the equation of the tangent line
The equation of the tangent line at point \( R \) can be expressed as:
\[ y - f(x) = f'(x)(x - x) \]
This simplifies to:
\[ y = f'(x)(x - x) + f(x) \]
Since the tangent intersects the y-axis at point \( P \), we can set \( x = 0 \) to find the y-intercept:
\[ y = f'(x)(0 - x) + f(x) \]
Thus, the y-coordinate of point \( P \) is:
\[ P = (0, f(x) - f'(x)x) \]

### Step 3: Determine the slope of the perpendicular line
The slope of the tangent line at point \( R \) is \( f'(x) \). Therefore, the slope of the line perpendicular to the tangent line is:
\[ -\frac{1}{f'(x)} \]

### Step 4: Equation of the perpendicular line through point \( P \)
The equation of the line perpendicular to the tangent through point \( P \) can be written as:
\[ y - (f(x) - f'(x)x) = -\frac{1}{f'(x)}(x - 0) \]
This simplifies to:
\[ y = -\frac{1}{f'(x)}x + f(x) - f'(x)x \]

### Step 5: Substitute the point (1, 0)
Since this line passes through the point \( (1, 0) \), we can substitute \( x = 1 \) and \( y = 0 \):
\[ 0 = -\frac{1}{f'(x)}(1) + f(x) - f'(x)x \]
This leads to:
\[ f(x) - f'(x)x = \frac{1}{f'(x)} \]

### Step 6: Rearranging the equation
Rearranging the equation gives us:
\[ f(x)f'(x) - f'(x)^2x = 1 \]

### Step 7: Change variables
Let \( y = f(x) \). Then, we can express the equation as:
\[ y f'(x) - (f'(x))^2 x = 1 \]

### Step 8: Formulate the differential equation
This can be rewritten in terms of derivatives:
\[ y \frac{dy}{dx} - x \left( \frac{dy}{dx} \right)^2 = 1 \]

### Conclusion
Thus, the differential equation of the curve is:
\[ y \frac{dy}{dx} - x \left( \frac{dy}{dx} \right)^2 = 1 \]

### Final Answer
The correct option is (a).

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