The curve with the property that the projection of the ordinate on the normal is constant and has a length equal to `a` is (a) `( b ) (c) a1n(( d ) (e)sqrt(( f ) (g) (h) y^(( i )2( j ))( k )-( l ) a^(( m )2( n ))( o ) (p))( q )+y (r))=x+c (s)` (t) (u) `( v ) (w) x+sqrt(( x ) (y) (z) a^(( a a )2( b b ))( c c )-( d d ) y^(( e e )2( f f ))( g g ) (hh))( i i )=c (jj)` (kk) (ll) `( m m ) (nn) (oo) (pp)(( q q ) (rr) y-a (ss))^(( t t )2( u u ))( v v )=c x (ww)` (xx) (yy) `( z z ) (aaa) a y=( b b b ) (ccc)tan^(( d d d ) (eee)-1( f f f ))( g g g )(( h h h ) (iii) x+c (jjj))( k k k )` (lll)

Options

A. `a" ln "(sqrt(y^(2)-a^(2)))=x+c`
B. `x+sqrt(a^(2)-y^(2))=c`
C. `(y-a)^(2)=cx`
D. `ay=tan^(-1)(x+c)`

Explanation

To solve the problem, we need to find the equation of the curve where the projection of the ordinate (y-coordinate) on the normal is constant and equal to a. Let's break down the solution step by step.

### Step 1: Understanding the Geometry
Let \( P(x, y) \) be a point on the curve. The normal at this point makes an angle \( \theta \) with the x-axis. The projection of the ordinate \( y \) onto the normal is given to be constant and equal to \( a \).

### Step 2: Setting Up the Relationship
From the geometry, we can express the projection of \( y \) onto the normal:
\[ y \cos \theta = a \]
This implies:
\[ \cos \theta = \frac{a}{y} \]

### Step 3: Using the Slope of the Normal
The slope of the normal line at point \( P \) is given by:
\[ \text{slope of normal} = -\frac{dx}{dy} \]
Thus, we have:
\[ \theta = \pi - \frac{dy}{dx} \implies \frac{dy}{dx} = \pi - \theta \]

### Step 4: Finding \( \sin \theta \)
Using the Pythagorean identity, we can find \( \sin \theta \):
\[ \sin^2 \theta + \cos^2 \theta = 1 \implies \sin^2 \theta = 1 - \left(\frac{a}{y}\right)^2 \]
Thus,
\[ \sin \theta = \sqrt{1 - \frac{a^2}{y^2}} = \frac{\sqrt{y^2 - a^2}}{y} \]

### Step 5: Relating \( \tan \theta \) to \( \frac{dy}{dx} \)
From the definition of tangent:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{y^2 - a^2}}{a} \]
Since \( \tan \theta = \frac{dy}{dx} \), we have:
\[ \frac{dy}{dx} = \frac{\sqrt{y^2 - a^2}}{a} \]

### Step 6: Separating Variables
We can separate the variables:
\[ a \, dy = \sqrt{y^2 - a^2} \, dx \]

### Step 7: Integrating Both Sides
Integrate both sides:
\[ \int \frac{a \, dy}{\sqrt{y^2 - a^2}} = \int dx \]
The left side integrates to:
\[ a \ln |y + \sqrt{y^2 - a^2}| + C_1 \]
The right side integrates to:
\[ x + C_2 \]

### Step 8: Combining Results
Setting the constants equal, we can write:
\[ x + C = a \ln |y + \sqrt{y^2 - a^2}| \]

### Final Equation
Rearranging gives us the final form of the equation:
\[ x + C = a \ln |y + \sqrt{y^2 - a^2}| \]