To solve the problem, we need to find the normal to the curve at the point \( P(0, -3) \) and determine where this normal intersects the curve again at two points where the tangents to the curve are the same.
### Step 1: Rewrite the curve equation
The given curve is:
\[
5x^5 - 10x^3 + x + 2y + 6 = 0
\]
We can rearrange this to express \( y \) in terms of \( x \):
\[
2y = -5x^5 + 10x^3 - x - 6 \implies y = -\frac{5}{2}x^5 + 5x^3 - \frac{1}{2}x - 3
\]
### Step 2: Find the derivative \( \frac{dy}{dx} \)
To find the slope of the tangent line at any point on the curve, we differentiate \( y \) with respect to \( x \):
\[
\frac{dy}{dx} = -\frac{25}{2}x^4 + 15x^2 - \frac{1}{2}
\]
### Step 3: Evaluate the derivative at \( P(0, -3) \)
Now we evaluate the derivative at the point \( P(0, -3) \):
\[
\frac{dy}{dx} \bigg|_{(0, -3)} = -\frac{25}{2}(0)^4 + 15(0)^2 - \frac{1}{2} = -\frac{1}{2}
\]
### Step 4: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[
\text{slope of normal} = -\frac{1}{\left(-\frac{1}{2}\right)} = 2
\]
### Step 5: Write the equation of the normal line
Using the point-slope form of the line equation, the equation of the normal line at \( P(0, -3) \) is:
\[
y - (-3) = 2(x - 0) \implies y + 3 = 2x \implies y = 2x - 3
\]
### Step 6: Substitute \( y \) into the curve equation
Next, we substitute \( y = 2x - 3 \) into the curve equation to find the points where the normal intersects the curve again:
\[
5x^5 - 10x^3 + x + 2(2x - 3) + 6 = 0
\]
Simplifying this gives:
\[
5x^5 - 10x^3 + x + 4x - 6 + 6 = 0 \implies 5x^5 - 10x^3 + 5x = 0
\]
Factoring out \( 5x \):
\[
5x(x^4 - 2x^2 + 1) = 0
\]
This gives us \( x = 0 \) or \( x^4 - 2x^2 + 1 = 0 \).
### Step 7: Solve the quartic equation
Let \( u = x^2 \), then we have:
\[
u^2 - 2u + 1 = 0 \implies (u - 1)^2 = 0 \implies u = 1 \implies x^2 = 1 \implies x = 1 \text{ or } x = -1
\]
### Step 8: Find corresponding \( y \) values
Now we find the \( y \) values for \( x = 1 \) and \( x = -1 \):
- For \( x = 1 \):
\[
y = 2(1) - 3 = -1 \implies (1, -1)
\]
- For \( x = -1 \):
\[
y = 2(-1) - 3 = -5 \implies (-1, -5)
\]
### Step 9: Verify the slopes of the tangents at points \( (1, -1) \) and \( (-1, -5) \)
Now we need to find the slopes of the tangents at these points:
1. At \( (1, -1) \):
\[
\frac{dy}{dx} \bigg|_{(1, -1)} = -\frac{25}{2}(1)^4 + 15(1)^2 - \frac{1}{2} = -\frac{25}{2} + 15 - \frac{1}{2} = 2
\]
2. At \( (-1, -5) \):
\[
\frac{dy}{dx} \bigg|_{(-1, -5)} = -\frac{25}{2}(-1)^4 + 15(-1)^2 - \frac{1}{2} = -\frac{25}{2} + 15 - \frac{1}{2} = 2
\]
Both points have the same slope of the tangent, confirming that the equation of the tangents at these points is the same.
### Conclusion
The normal at point \( P(0, -3) \) intersects the curve again at points \( (1, -1) \) and \( (-1, -5) \), where the equations of the tangents are the same.
