Let A be the point where the curve `5 alpha ^(2) x ^(3) +10 alpha x ^(2)+ x + 2y - 4=0 (alpha in R, alpha ne 0)` meets the y-axis, then the equation of tangent to the curve at the point where normal at A meets the curve again, is:

Options

A. `x-alphay+2 alpha=0`
B. `alpha x+ y -2 =0`
C. `2x -y+2 =0`
D. `x+2y-4=0`

Answer

To solve the given problem, we will follow these steps:

### Step 1: Find the point A where the curve meets the y-axis.
The curve is given by the equation:
\[ 5 \alpha^2 x^3 + 10 \alpha x^2 + x + 2y - 4 = 0 \]

At the y-axis, \( x = 0 \). Substituting \( x = 0 \) into the equation:
\[ 5 \alpha^2 (0)^3 + 10 \alpha (0)^2 + (0) + 2y - 4 = 0 \]
This simplifies to:
\[ 2y - 4 = 0 \]
Thus,
\[ 2y = 4 \]
\[ y = 2 \]

So, the coordinates of point A are:
\[ A(0, 2) \]

### Step 2: Find the slope of the tangent at point A.
To find the slope of the tangent, we need to differentiate the curve with respect to \( x \).
Starting with the equation:
\[ 5 \alpha^2 x^3 + 10 \alpha x^2 + x + 2y - 4 = 0 \]

Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(5 \alpha^2 x^3) + \frac{d}{dx}(10 \alpha x^2) + \frac{d}{dx}(x) + \frac{d}{dx}(2y) = 0 \]

This gives:
\[ 15 \alpha^2 x^2 + 20 \alpha x + 1 + 2 \frac{dy}{dx} = 0 \]

Rearranging for \( \frac{dy}{dx} \):
\[ 2 \frac{dy}{dx} = - (15 \alpha^2 x^2 + 20 \alpha x + 1) \]
\[ \frac{dy}{dx} = -\frac{1}{2} (15 \alpha^2 x^2 + 20 \alpha x + 1) \]

Now substituting \( x = 0 \) to find the slope at point A:
\[ \frac{dy}{dx} \bigg|_{(0, 2)} = -\frac{1}{2} (15 \alpha^2 (0)^2 + 20 \alpha (0) + 1) = -\frac{1}{2} \]

Thus, the slope of the tangent at point A is:
\[ m_t = -\frac{1}{2} \]

### Step 3: Find the slope of the normal at point A.
The slope of the normal is the negative reciprocal of the slope of the tangent:
\[ m_n = -\frac{1}{m_t} = -\frac{1}{-\frac{1}{2}} = 2 \]

### Step 4: Write the equation of the normal at point A.
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (0, 2) \) and \( m = 2 \):
\[ y - 2 = 2(x - 0) \]
This simplifies to:
\[ y = 2x + 2 \]

### Step 5: Find the intersection of the normal with the curve again.
We substitute \( y = 2x + 2 \) into the original curve equation:
\[ 5 \alpha^2 x^3 + 10 \alpha x^2 + x + 2(2x + 2) - 4 = 0 \]
This simplifies to:
\[ 5 \alpha^2 x^3 + 10 \alpha x^2 + x + 4x + 4 - 4 = 0 \]
\[ 5 \alpha^2 x^3 + 10 \alpha x^2 + 5x = 0 \]
Factoring out \( 5x \):
\[ 5x(\alpha^2 x^2 + 2\alpha x + 1) = 0 \]

This gives us one solution \( x = 0 \) (point A). The other solution comes from:
\[ \alpha^2 x^2 + 2\alpha x + 1 = 0 \]

Using the quadratic formula:
\[ x = \frac{-2\alpha \pm \sqrt{(2\alpha)^2 - 4\alpha^2}}{2\alpha^2} = \frac{-2\alpha \pm 0}{2\alpha^2} = -\frac{1}{\alpha} \]

### Step 6: Find the coordinates of the point Q.
Substituting \( x = -\frac{1}{\alpha} \) into \( y = 2x + 2 \):
\[ y = 2\left(-\frac{1}{\alpha}\right) + 2 = -\frac{2}{\alpha} + 2 = 2 - \frac{2}{\alpha} \]

Thus, the coordinates of point Q are:
\[ Q\left(-\frac{1}{\alpha}, 2 - \frac{2}{\alpha}\right) \]

### Step 7: Find the slope of the tangent at point Q.
Using the slope formula:
\[ m_t = -\frac{1}{2}(15\alpha^2(-\frac{1}{\alpha})^2 + 20\alpha(-\frac{1}{\alpha}) + 1) \]
This simplifies to:
\[ m_t = -\frac{1}{2}(15 + (-20) + 1) = -\frac{1}{2}(-4) = 2 \]

### Step 8: Write the equation of the tangent at point Q.
Using the point-slope form:
\[ y - \left(2 - \frac{2}{\alpha}\right) = 2\left(x + \frac{1}{\alpha}\right) \]
This simplifies to:
\[ y = 2x + 2 \]

### Final Answer:
The equation of the tangent to the curve at the point where the normal at A meets the curve again is:
\[ y = 2x + 2 \]