The co-ordinates of the points on the curve `y=x^(2)+3x+4` at which the tangent passes through the origin are (A) `(-2,14)` (B) `(2,14)` (C) `(2,-2)` (D) `(-2,2)`

Options

A. `(-2,14)`
B. `(2,14)`
C. `(2,-2)`
D. `(-2,2)`

Explanation

To find the coordinates of the points on the curve \( y = x^2 + 3x + 4 \) at which the tangent passes through the origin, we can follow these steps:

### Step 1: Find the derivative of the function
The first step is to find the derivative of the function to determine the slope of the tangent line at any point on the curve.

\[ y = x^2 + 3x + 4 \]

Taking the derivative with respect to \( x \):

\[ \frac{dy}{dx} = 2x + 3 \]

### Step 2: Set up the equation of the tangent line
Let’s assume the point of tangency is \( (x_1, y_1) \). The coordinates \( y_1 \) can be expressed as:

\[ y_1 = x_1^2 + 3x_1 + 4 \]

The slope of the tangent line at this point is:

\[ m = 2x_1 + 3 \]

The equation of the tangent line at the point \( (x_1, y_1) \) can be expressed using point-slope form:

\[ y - y_1 = m(x - x_1) \]

Substituting for \( m \) and \( y_1 \):

\[ y - (x_1^2 + 3x_1 + 4) = (2x_1 + 3)(x - x_1) \]

### Step 3: Determine when the tangent passes through the origin
For the tangent line to pass through the origin, we set \( x = 0 \) and \( y = 0 \):

\[ 0 - (x_1^2 + 3x_1 + 4) = (2x_1 + 3)(0 - x_1) \]

This simplifies to:

\[ -(x_1^2 + 3x_1 + 4) = -(2x_1 + 3)x_1 \]

### Step 4: Simplify the equation
Removing the negative signs gives:

\[ x_1^2 + 3x_1 + 4 = (2x_1 + 3)x_1 \]

Expanding the right side:

\[ x_1^2 + 3x_1 + 4 = 2x_1^2 + 3x_1 \]

### Step 5: Rearranging the equation
Now, rearranging the equation:

\[ x_1^2 + 3x_1 + 4 - 2x_1^2 - 3x_1 = 0 \]

This simplifies to:

\[ -x_1^2 + 4 = 0 \]

### Step 6: Solve for \( x_1 \)
Rearranging gives:

\[ x_1^2 = 4 \]

Taking the square root of both sides:

\[ x_1 = 2 \quad \text{or} \quad x_1 = -2 \]

### Step 7: Find corresponding \( y_1 \) values
Now we need to find the corresponding \( y_1 \) values for both \( x_1 \):

1. For \( x_1 = 2 \):

\[ y_1 = 2^2 + 3(2) + 4 = 4 + 6 + 4 = 14 \]

So one point is \( (2, 14) \).

2. For \( x_1 = -2 \):

\[ y_1 = (-2)^2 + 3(-2) + 4 = 4 - 6 + 4 = 2 \]

So the other point is \( (-2, 2) \).

### Final Answer
The coordinates of the points on the curve at which the tangent passes through the origin are:

- \( (2, 14) \)
- \( (-2, 2) \)