To solve the problem, we need to analyze the two statements provided regarding the parabola \( y^2 = 4ax \) and the point \( (a, a+1) \).
### Step 1: Determine the position of the point \( (a, a+1) \) with respect to the parabola.
To find the position of the point relative to the parabola, we will use the condition derived from the equation of the parabola. The equation of the parabola is given by:
\[
y^2 - 4ax = 0
\]
We can substitute the coordinates of the point \( (a, a+1) \) into this equation to find \( S \):
\[
S = (a+1)^2 - 4a(a)
\]
### Step 2: Simplify the expression for \( S \).
Expanding \( S \):
\[
S = (a^2 + 2a + 1) - 4a^2
\]
\[
S = a^2 + 2a + 1 - 4a^2
\]
\[
S = -3a^2 + 2a + 1
\]
### Step 3: Analyze the sign of \( S \).
To determine whether the point lies inside, on, or outside the parabola, we analyze the sign of \( S \):
- If \( S > 0 \), the point lies outside the parabola.
- If \( S = 0 \), the point lies on the parabola.
- If \( S < 0 \), the point lies inside the parabola.
### Step 4: Find the roots of the quadratic equation \( -3a^2 + 2a + 1 = 0 \).
To find the roots, we can use the quadratic formula:
\[
a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = -3, b = 2, c = 1 \):
\[
a = \frac{-2 \pm \sqrt{2^2 - 4(-3)(1)}}{2(-3)}
\]
\[
= \frac{-2 \pm \sqrt{4 + 12}}{-6}
\]
\[
= \frac{-2 \pm \sqrt{16}}{-6}
\]
\[
= \frac{-2 \pm 4}{-6}
\]
Calculating the roots:
1. \( a_1 = \frac{2}{-6} = -\frac{1}{3} \)
2. \( a_2 = \frac{-6}{-6} = 1 \)
### Step 5: Determine the intervals for \( S \).
The roots divide the number line into intervals. We can test the sign of \( S \) in the intervals:
1. For \( a < -\frac{1}{3} \): Choose \( a = -1 \):
\[
S = -3(-1)^2 + 2(-1) + 1 = -3 - 2 + 1 = -4 \quad (\text{negative})
\]
2. For \( -\frac{1}{3} < a < 1 \): Choose \( a = 0 \):
\[
S = -3(0)^2 + 2(0) + 1 = 1 \quad (\text{positive})
\]
3. For \( a > 1 \): Choose \( a = 2 \):
\[
S = -3(2)^2 + 2(2) + 1 = -12 + 4 + 1 = -7 \quad (\text{negative})
\]
### Conclusion for Statement 1:
From our analysis:
- \( S > 0 \) for \( -\frac{1}{3} < a < 1 \) (point lies outside).
- \( S < 0 \) for \( a < -\frac{1}{3} \) and \( a > 1 \) (point lies inside).
Thus, the statement that "three normals can be drawn to the parabola through the point \( (a, a+1) \) if \( a < 2 \)" is **false** because it is not guaranteed that three normals can be drawn for all \( a < 2 \).
### Conclusion for Statement 2:
The point \( (a, a+1) \) lies outside the parabola for all \( a \neq 1 \) as we derived that \( S > 0 \) for \( a \) not equal to 1. Therefore, Statement 2 is **true**.
### Final Answer:
- Statement 1 is false.
- Statement 2 is true.
