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. `x(d^(2)y)/(dx^(2))+((dy)/(dx))^(2)=1`
C. `y.(dx)/(dy)+x = 1`
D. None of these

Explanation

To solve the problem step by step, we will derive the differential equation of the curve based on the given conditions.

### Step 1: Understand the Tangent Line
The equation of the tangent line to the curve \( y = f(x) \) at the point \( (x, f(x)) \) can be expressed as:
\[ y - f(x) = f'(x)(x - x_0) \]
where \( f'(x) \) is the derivative of \( f(x) \) at point \( x_0 \).

### Step 2: Find the y-intercept
The y-intercept occurs when \( x = 0 \). Substituting \( x = 0 \) into the tangent equation gives:
\[ y - f(x) = f'(x)(0 - x_0) \implies y = f(x) - f'(x)x_0 \]
Let \( P \) be the point where the tangent intercepts the y-axis, so the coordinates of \( P \) are \( (0, f(x) - f'(x)x_0) \).

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

### Step 4: Equation of the Perpendicular Line
The equation of the line perpendicular to the tangent through point \( P(0, f(x) - f'(x)x_0) \) can be written as:
\[ y - (f(x) - f'(x)x_0) = -\frac{1}{f'(x)}(x - 0) \]
This simplifies to:
\[ y = -\frac{1}{f'(x)}x + \left(f(x) - f'(x)x_0\right) \]

### Step 5: Condition of Passing Through (1, 0)
Since this perpendicular line passes through the point \( (1, 0) \), we can substitute \( x = 1 \) and \( y = 0 \):
\[ 0 = -\frac{1}{f'(x)}(1) + \left(f(x) - f'(x)x_0\right) \]
Rearranging gives:
\[ f'(x)x_0 = f(x) - \frac{1}{f'(x)} \]

### Step 6: Deriving the Differential Equation
From the equation above, we can express \( x_0 \) in terms of \( f(x) \) and \( f'(x) \):
\[ x_0 = \frac{f(x) - \frac{1}{f'(x)}}{f'(x)} \]
Now, substituting \( x_0 \) back into the tangent equation and simplifying leads us to the differential equation:
\[ y \frac{dy}{dx} - x \left(\frac{dy}{dx}\right)^2 = 1 \]

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

### Conclusion
The correct option for the differential equation of the curve is (a).

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