A tangent `(x_1,y_1)` to the curve `y=x^3+2x^2+4` and passes through origin then `(x_1,y_1)` is

Options

A. (0,4)
B. (-1,5)
C. (1,7)
D. (2,20)

Answer

To solve the problem, we need to find the point \((x_1, y_1)\) on the curve \(y = x^3 + 2x^2 + 4\) where the tangent at that point passes through the origin \((0, 0)\).

### Step-by-step Solution:

1. **Differentiate the Curve**:
We start by finding the derivative of the curve to determine the slope of the tangent line at any point \(x_1\).
\[ \frac{dy}{dx} = 3x^2 + 4x \]

2. **Evaluate the Slope at \(x_1\)**:
The slope of the tangent line at the point \((x_1, y_1)\) is:
\[ m = 3x_1^2 + 4x_1 \]

3. **Equation of the Tangent Line**:
The equation of the tangent line at the point \((x_1, y_1)\) can be written using the point-slope form:
\[ y - y_1 = m(x - x_1) \]
Since the tangent passes through the origin \((0, 0)\), we can substitute these values into the equation:
\[ 0 - y_1 = (3x_1^2 + 4x_1)(0 - x_1) \]
Simplifying this gives:
\[ -y_1 = -(3x_1^2 + 4x_1)x_1 \]
\[ y_1 = (3x_1^2 + 4x_1)x_1 = 3x_1^3 + 4x_1^2 \]

4. **Substituting \(y_1\) into the Curve Equation**:
The point \((x_1, y_1)\) lies on the curve, so we substitute \(y_1\) into the curve equation:
\[ y_1 = x_1^3 + 2x_1^2 + 4 \]
Setting the two expressions for \(y_1\) equal gives:
\[ 3x_1^3 + 4x_1^2 = x_1^3 + 2x_1^2 + 4 \]

5. **Rearranging the Equation**:
Rearranging this equation leads to:
\[ 3x_1^3 + 4x_1^2 - x_1^3 - 2x_1^2 - 4 = 0 \]
\[ 2x_1^3 + 2x_1^2 - 4 = 0 \]
Dividing the entire equation by 2:
\[ x_1^3 + x_1^2 - 2 = 0 \]

6. **Finding Roots of the Polynomial**:
We can try \(x_1 = 1\):
\[ 1^3 + 1^2 - 2 = 1 + 1 - 2 = 0 \]
Thus, \(x_1 = 1\) is a root.

7. **Finding \(y_1\)**:
Now substituting \(x_1 = 1\) back into the expression for \(y_1\):
\[ y_1 = 3(1)^3 + 4(1)^2 = 3 + 4 = 7 \]

8. **Final Point**:
Therefore, the point \((x_1, y_1)\) is:
\[ (1, 7) \]

### Conclusion:
The point \((x_1, y_1)\) where the tangent to the curve passes through the origin is \((1, 7)\).