The normal at any point `P(x_1,y_1)` of curve is a line perpendicular to tangent at the point `P(x_1,y_1)`. In case of rectangular hyperbola `xy=c^2`, the equation of normal at `(ct,(c )/(t))` is `xt^3-yt-ct^4+c=0`. The shortest distance between any two curves always along the common normal.
The number of normals drawn from `((7)/(6),4)` to parabola `y^2=2x-1` is :

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the number of normals drawn from the point \((\frac{7}{6}, 4)\) to the parabola given by the equation \(y^2 = 2x - 1\), we will follow these steps:

### Step 1: Understand the parabola
The equation of the parabola can be rewritten as:
\[ y^2 = 2x - 1 \]
This is a right-handed opening parabola.

### Step 2: Determine the position of the point relative to the parabola
To find out if the point \((\frac{7}{6}, 4)\) lies inside or outside the parabola, we can substitute \(x = \frac{7}{6}\) and \(y = 4\) into the equation of the parabola.

Substituting into the equation:
\[ 4^2 = 2\left(\frac{7}{6}\right) - 1 \]
Calculating the left side:
\[ 16 = 2 \cdot \frac{7}{6} - 1 \]
Calculating the right side:
\[ 16 = \frac{14}{6} - 1 = \frac{14}{6} - \frac{6}{6} = \frac{8}{6} = \frac{4}{3} \]

Since \(16 > \frac{4}{3}\), the point \((\frac{7}{6}, 4)\) lies outside the parabola.

### Step 3: Determine the number of normals
For a point outside a parabola, there can be one or two normals drawn to the parabola. To find the exact number, we can analyze the situation geometrically:

1. **If the point lies outside the parabola**, it can have:
- One normal if it is tangent to the parabola at one point.
- Two normals if it intersects the parabola at two distinct points.

Since we have established that the point \((\frac{7}{6}, 4)\) is outside the parabola, we need to check if it can have two normals.

### Step 4: Check for tangents
To find the normals from the point to the parabola, we can derive the equations for the normals. The general form of the normal to the parabola \(y^2 = 2x - 1\) at a point \((x_0, y_0)\) on the parabola is given by:
\[ y - y_0 = -\frac{y_0}{(y_0 + 1)}(x - x_0) \]

However, since we are looking for the number of normals, we can conclude:
- The point \((\frac{7}{6}, 4)\) being outside the parabola indicates that there can only be one normal line that can be drawn from this point to the parabola.

### Conclusion
Thus, the number of normals drawn from the point \((\frac{7}{6}, 4)\) to the parabola \(y^2 = 2x - 1\) is:
\[ \boxed{1} \]