Question

If \(x + 13y = 40\) is normal to the curve \(y = 5x^{2} + \alpha x + \beta\) at the point (1,3), then the value of \(\alpha\beta\) is equal to:

• A. 15
• B. -6
• C. 6
• D. 13
• E. -15

Hint:

Always check if the given point actually lies on the given line first. Here, \(1 + 13(3) = 1 + 39 = 40\), which confirms (1, 3) is the correct point of contact. This prevents errors from misread coordinates.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
A line is normal to a curve at a point if its slope is the negative reciprocal of the tangent's slope at that point. Additionally, the point of tangency must satisfy both the equation of the curve and the equation of the line.

Step 2: Key Formula or Approach:
1. Slope of normal (\(m_n\)) \(\times\) Slope of tangent (\(y'\)) = \(-1\).
2. Point (1, 3) must satisfy \(y = 5x^2 + \alpha x + \beta\).

Step 3: Detailed Explanation:
1. Find slope of the normal: From \(x + 13y = 40\), \(13y = -x + 40 \implies y = -\frac{1}{13}x + \frac{40}{13}\). The slope of the normal \(m_n = -1/13\).
2. Find slope of the tangent: Since \(m_n \cdot m_t = -1\), the tangent slope \(m_t = 13\).
3. Relate to curve derivative: \(y' = 10x + \alpha\). At \(x=1\), \(y' = 10(1) + \alpha = 13\). \[ 10 + \alpha = 13 \implies \alpha = 3 \]
4. Use the point (1, 3) on the curve: \[ 3 = 5(1)^2 + \alpha(1) + \beta \] \[ 3 = 5 + 3 + \beta \implies 3 = 8 + \beta \implies \beta = -5 \]
5. Calculate \(\alpha\beta\): \(3 \times (-5) = -15\).

Step 4: Final Answer
The value of \(\alpha\beta\) is -15.