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:
The problem relates a given curve to its normal line at a specific point of contact.
The normal line is always perpendicular to the tangent line at the point of tangency.
Therefore, the slope of the normal and the slope of the tangent are negative reciprocals of each other.
We can find the tangent's slope using the derivative of the curve's equation.
Step 2: Key Formula or Approach:
1. Point condition: If a line is normal/tangent to a curve at point \( (x_1, y_1) \), that point must satisfy the equation of the curve \( y = f(x) \).
2. Slope of Normal (\(m_n\)): Extracted from the linear equation \( y = m_n x + c \).
3. Slope of Tangent (\(m_t\)): Given by \( m_t = \left. \frac{dy}{dx} \right|_{(x_1, y_1)} \).
4. Perpendicularity condition: \( m_t \times m_n = -1 \), which implies \( m_t = -\frac{1}{m_n} \).
Step 3: Detailed Explanation:
First, since the normal is at the point \( (1,3) \), this point lies exactly on the given curve \( y = 5x^2 + ax + \beta \).
Substitute \( x = 1 \) and \( y = 3 \) into the curve's equation to create our first relation:
\[ 3 = 5(1)^2 + a(1) + \beta \] \[ 3 = 5 + a + \beta \] \[ a + \beta = 3 - 5 \] \[ a + \beta = -2 \quad \text{--- (Equation 1)} \] Next, let's find the slope of the normal line. The equation is given as \( x + 13y = 40 \).
Rearrange it into the slope-intercept form (\( y = mx + c \)):
\[ 13y = -x + 40 \] \[ y = -\frac{1}{13}x + \frac{40}{13} \] From this, the slope of the normal line is \( m_n = -\frac{1}{13} \).
Using the perpendicularity condition, the slope of the tangent line \( m_t \) is the negative reciprocal:
\[ m_t = -\frac{1}{m_n} = -\frac{1}{-\frac{1}{13}} = 13 \] Now, we find the slope of the tangent analytically by taking the derivative of the curve \( y = 5x^2 + ax + \beta \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(5x^2 + ax + \beta) = 10x + a \] Evaluate this derivative specifically at the point of tangency, where \( x = 1 \):
\[ \left. \frac{dy}{dx} \right|_{x=1} = 10(1) + a = 10 + a \] Equate this evaluated derivative to the tangent slope \( m_t \) we found from the normal line:
\[ 10 + a = 13 \] Solving for \( a \):
\[ a = 3 \] Now, substitute the value of \( a = 3 \) back into Equation 1 to solve for \( \beta \):
\[ 3 + \beta = -2 \] \[ \beta = -2 - 3 \] \[ \beta = -5 \] The problem asks for the product of \( a \) and \( \beta \):
\[ a\beta = (3)(-5) = -15 \] Step 4: Final Answer:
The product \( a\beta \) is equal to -15.