The normal to the curve `5x^(5)-10x^(3)+x+2y+6 =0` at P(0, -3) meets the curve again at the point

Options

A. (1,-1),(-1,-2)
B. (1,2),(-1,3)
C. (1,-1),(-1,5)
D. (1,-3),(-1,-2)

Answer

To solve the problem, we will follow these steps:

### Step 1: Differentiate the curve equation
We start with the curve given by the equation:
\[ 5x^5 - 10x^3 + x + 2y + 6 = 0 \]

To find the slope of the tangent at the point \( P(0, -3) \), we differentiate the equation implicitly with respect to \( x \):
\[ \frac{d}{dx}(5x^5) - \frac{d}{dx}(10x^3) + \frac{d}{dx}(x) + 2\frac{dy}{dx} + 0 = 0 \]
This gives:
\[ 25x^4 - 30x^2 + 1 + 2\frac{dy}{dx} = 0 \]

### Step 2: Find the slope of the tangent at point P
Now, we substitute \( x = 0 \) into the differentiated equation to find \( \frac{dy}{dx} \) at the point \( P(0, -3) \):
\[ 25(0)^4 - 30(0)^2 + 1 + 2\frac{dy}{dx} = 0 \]
This simplifies to:
\[ 1 + 2\frac{dy}{dx} = 0 \implies 2\frac{dy}{dx} = -1 \implies \frac{dy}{dx} = -\frac{1}{2} \]

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent. Thus:
\[ \text{slope of normal} = -\frac{1}{\left(-\frac{1}{2}\right)} = 2 \]

### Step 4: Write the equation of the normal line
Using the point-slope form of the line equation, the equation of the normal line at point \( P(0, -3) \) is:
\[ y - (-3) = 2(x - 0) \implies y + 3 = 2x \implies y = 2x - 3 \]

### Step 5: Substitute the normal line equation into the curve equation
To find where this normal line intersects the curve again, we substitute \( y = 2x - 3 \) into the original curve equation:
\[ 5x^5 - 10x^3 + x + 2(2x - 3) + 6 = 0 \]
This simplifies to:
\[ 5x^5 - 10x^3 + x + 4x - 6 + 6 = 0 \implies 5x^5 - 10x^3 + 5x = 0 \]

### Step 6: Factor the equation
Factoring out \( 5x \):
\[ 5x(x^4 - 2x^2 + 1) = 0 \]
This gives us:
\[ x = 0 \quad \text{or} \quad x^4 - 2x^2 + 1 = 0 \]

### Step 7: Solve the quadratic in terms of \( x^2 \)
Let \( u = x^2 \):
\[ u^2 - 2u + 1 = 0 \implies (u - 1)^2 = 0 \implies u = 1 \implies x^2 = 1 \implies x = \pm 1 \]

### Step 8: Find the corresponding \( y \) values
Now we find the corresponding \( y \) values for \( x = 1 \) and \( x = -1 \):
1. For \( x = 1 \):
\[ 5(1)^5 - 10(1)^3 + 1 + 2y + 6 = 0 \implies 5 - 10 + 1 + 2y + 6 = 0 \implies 2y + 2 = 0 \implies y = -1 \]
So, the point is \( (1, -1) \).

2. For \( x = -1 \):
\[ 5(-1)^5 - 10(-1)^3 + (-1) + 2y + 6 = 0 \implies -5 + 10 - 1 + 2y + 6 = 0 \implies 2y + 10 = 0 \implies y = -5 \]
So, the point is \( (-1, -5) \).

### Conclusion
The normal to the curve at \( P(0, -3) \) meets the curve again at the points \( (1, -1) \) and \( (-1, -5) \).