Question

If the points of contact of the tangents drawn from \( (0,0) \) to the curve \( y = x^2 + 3x + 4 \) are \( (\alpha, \beta) \) and \( (\gamma, \delta) \), then \( \beta + \delta = \):

• A. 7
• B. 25
• C. 16
• D. 13

Hint:

When dealing with tangents to curves, use the derivative to find the slope of the tangent, and apply the point-slope form of the line.

Answer 1

The Correct Option is C

We are given the curve \( y = x^2 + 3x + 4 \) and the point \( (0, 0) \) from which the tangents are drawn. The points of contact of the tangents with the curve are \( (\alpha, \beta) \) and \( (\gamma, \delta) \). Step 1: Equation of the tangent to the curve The equation of the tangent to the curve \( y = x^2 + 3x + 4 \) at any point \( (x_1, y_1) \) is given by: \[ y - y_1 = f'(x_1)(x - x_1) \] where \( f'(x_1) \) is the derivative of \( y = x^2 + 3x + 4 \). Differentiating: \[ f'(x) = 2x + 3 \] Substitute \( y_1 = x_1^2 + 3x_1 + 4 \) and \( f'(x_1) = 2x_1 + 3 \) into the tangent equation: \[ y - (x_1^2 + 3x_1 + 4) = (2x_1 + 3)(x - x_1) \] Step 2: Finding the points of contact The tangents from the origin \( (0, 0) \) will satisfy the equation of the tangent. We use the condition that the distance from the origin to the tangent line is zero. After solving, we find that \( \beta + \delta = 16 \). Thus, the correct answer is option (3), \( \beta + \delta = 16 \).

Answer 2

Step 1: Equation of tangent to the curve at point \((x_1, y_1)\)
Given curve: \( y = x^2 + 3x + 4 \)
Slope at \( x = x_1 \) is:
\[ m = \frac{dy}{dx} = 2x_1 + 3 \]
Equation of tangent at \((x_1, y_1)\) is:
\[ y - y_1 = m (x - x_1) \] where \( y_1 = x_1^2 + 3x_1 + 4 \).

Step 2: Condition for tangent passing through \((0,0)\)
Substitute \((0,0)\) in tangent equation:
\[ 0 - y_1 = m (0 - x_1) \implies -y_1 = -m x_1 \implies y_1 = m x_1 \]
Substitute \( y_1 \) and \( m \):
\[ x_1^2 + 3x_1 + 4 = (2x_1 + 3) x_1 = 2x_1^2 + 3x_1 \]

Step 3: Simplify the equation
\[ x_1^2 + 3x_1 + 4 = 2x_1^2 + 3x_1 \implies 4 = 2x_1^2 - x_1^2 = x_1^2 \]
Thus:
\[ x_1^2 = 4 \implies x_1 = \pm 2 \]

Step 4: Find the corresponding \( y \)-coordinates
\[ \beta = y(\alpha) = (2)^2 + 3 \cdot 2 + 4 = 4 + 6 + 4 = 14 \]
\[ \delta = y(\gamma) = (-2)^2 + 3 \cdot (-2) + 4 = 4 - 6 + 4 = 2 \]

Step 5: Calculate \(\beta + \delta\)
\[ \beta + \delta = 14 + 2 = 16 \]