To solve the problem step by step, we will analyze the conditions given in the question and derive the necessary equations.
### Step 1: Understand the Curve and the Normals
The given curve is \( y^2 = x \). We need to find the normals drawn from the point \( (c, 0) \) to this curve.
### Step 2: Find the Slope of the Normal
For the curve \( y^2 = x \), we can express \( x \) in terms of a parameter \( t \):
- Let \( x = t^2 \) and \( y = t \).
The derivative \( \frac{dy}{dx} \) can be found using implicit differentiation:
- From \( y^2 = x \), we have \( 2y \frac{dy}{dx} = 1 \) or \( \frac{dy}{dx} = \frac{1}{2y} \).
At the point \( (t^2, t) \), the slope of the tangent line is:
- \( \frac{dy}{dx} = \frac{1}{2t} \).
Thus, the slope of the normal line (which is the negative reciprocal) is:
- \( m = -2t \).
### Step 3: Equation of the Normal
Using the point-slope form, the equation of the normal line at the point \( (t^2, t) \) is:
\[
y - t = -2t(x - t^2)
\]
Rearranging gives:
\[
y = -2tx + 2t^3 + t.
\]
### Step 4: Substitute the Point (c, 0)
We need to find the normals that pass through the point \( (c, 0) \):
\[
0 = -2tc + 2t^3 + t.
\]
Rearranging this gives:
\[
2t^3 + t - 2tc = 0.
\]
Factoring out \( t \):
\[
t(2t^2 + 1 - 2c) = 0.
\]
This gives us one solution \( t = 0 \) (which corresponds to the normal along the x-axis). For the other normals, we need to solve:
\[
2t^2 + 1 - 2c = 0.
\]
### Step 5: Condition for Real Normals
For \( t \) to have real solutions, the discriminant of the quadratic equation must be non-negative:
\[
D = 1 - 4 \cdot 2 \cdot (-2c) = 1 + 16c \geq 0.
\]
This is always true for \( c \geq -\frac{1}{16} \). However, we also need to ensure that the quadratic has two distinct solutions.
### Step 6: Ensure Two Distinct Normals
To ensure two distinct normals, we need:
\[
1 - 8c > 0 \implies c < \frac{1}{8}.
\]
However, we also need to check the condition derived from the normal equation:
\[
2t^2 + 1 - 2c = 0 \implies 2c - 1 > 0 \implies c > \frac{1}{2}.
\]
### Conclusion for c
Thus, we conclude that \( c \) must be greater than \( \frac{1}{2} \) for there to be three normals from the point \( (c, 0) \) to the curve \( y^2 = x \).
### Step 7: Find c for Perpendicular Normals
For the two normals to be perpendicular, the product of their slopes must be -1:
- The slopes from the quadratic \( 2t^2 + 1 - 2c = 0 \) are given by the roots of the equation \( m^2 + 2 - 4c = 0 \).
Setting the product of the roots equal to -1:
\[
(2 - 4c) = -1 \implies 2 - 4c = -1 \implies 4c = 3 \implies c = \frac{3}{4}.
\]
### Final Result
1. \( c > \frac{1}{2} \) for three normals.
2. \( c = \frac{3}{4} \) for the two normals to be perpendicular.
