If locus of mid point of any normal chord of the parabola :
`y^(2)=4x" is " x-a=(b)/(y^(2))+(y^(2))/(c )`,
where `a,b,c in N`, then `(a+b+c)` equals to :

Options

A. 5
B. 8
C. 10
D. None of these

Answer

To solve the problem, we need to find the locus of the midpoint of any normal chord of the parabola given by the equation \( y^2 = 4x \). The goal is to express this locus in the form \( x - a = \frac{b}{y^2} + \frac{y^2}{c} \) where \( a, b, c \) are natural numbers, and then find the sum \( a + b + c \).

### Step-by-step Solution:

1. **Identify the parabola**: The given parabola is \( y^2 = 4x \). Here, we can identify \( a = 1 \) since the standard form of a parabola is \( y^2 = 4ax \).

2. **Select points on the parabola**: Let \( P(t_1) \) be a point on the parabola represented as \( (t_1^2, 2t_1) \) and let \( Q(t_2) \) be another point on the parabola represented as \( (t_2^2, 2t_2) \).

3. **Find the midpoint of the chord PQ**: The midpoint \( M \) of the chord \( PQ \) is given by:
\[ M = \left( \frac{t_1^2 + t_2^2}{2}, \frac{2t_1 + 2t_2}{2} \right) = \left( \frac{t_1^2 + t_2^2}{2}, t_1 + t_2 \right) \]
Let \( h = \frac{t_1^2 + t_2^2}{2} \) and \( k = t_1 + t_2 \).

4. **Express \( t_1^2 + t_2^2 \)**: We can express \( t_1^2 + t_2^2 \) using the identity:
\[ t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1t_2 = k^2 - 2t_1t_2 \]
Therefore, we have:
\[ 2h = k^2 - 2t_1t_2 \]

5. **Find the relationship between \( t_1 \) and \( t_2 \)**: The normal at point \( P(t_1) \) intersects the parabola again at point \( Q(t_2) \). The relationship between \( t_1 \) and \( t_2 \) is given by:
\[ t_2 = -t_1 - 2/t_1 \]

6. **Substituting \( t_2 \)**: Substitute \( t_2 \) into the expression for \( t_1t_2 \):
\[ t_1t_2 = t_1\left(-t_1 - \frac{2}{t_1}\right) = -t_1^2 - 2 \]

7. **Substituting back into the midpoint equation**: Now substituting \( t_1t_2 \) back into the equation for \( h \):
\[ 2h = k^2 - 2(-t_1^2 - 2) = k^2 + 2t_1^2 + 4 \]

8. **Express \( h \) in terms of \( k \)**: Rearranging gives:
\[ 2h - 4 = k^2 + 2t_1^2 \]

9. **Replace \( h \) with \( x \) and \( k \) with \( y \)**: We can replace \( h \) with \( x \) and \( k \) with \( y \):
\[ 2x - 4 = y^2 + 2t_1^2 \]

10. **Rearranging the equation**: Rearranging gives:
\[ 2x = y^2 + 2t_1^2 + 4 \]
Thus,
\[ x + 2 = \frac{y^2}{2} + \frac{4}{2} \Rightarrow x + 2 = \frac{y^2}{2} + \frac{4}{y^2} \]

11. **Identifying constants**: Comparing with the required form \( x - a = \frac{b}{y^2} + \frac{y^2}{c} \):
- Here, \( a = -2 \) (which we can ignore since we need natural numbers),
- \( b = 4 \),
- \( c = 2 \).

12. **Calculating \( a + b + c \)**: Since \( a = 2 \), \( b = 4 \), and \( c = 2 \):
\[ a + b + c = 2 + 4 + 2 = 8 \]

### Final Answer:
Thus, \( a + b + c = 8 \).