Question

If x+13y=40 is normal to the curve y=5x²+ax+β at the point (1, 3), then the value of αß is equal to

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

Answer 1

Answer: (E)

We are given the equation of the curve \( y = 5x^2 + \alpha x + \beta \) and the equation of the normal line \( x + 13y = 40 \), which is normal to the curve at the point \( (1, 3) \).
The equation of the normal line is \( x + 13y = 40 \). Rewriting this in slope-intercept form: \[ 13y = -x + 40 \quad \Rightarrow \quad y = -\frac{1}{13}x + \frac{40}{13}. \] Thus, the slope of the normal line is \( m_{\text{normal}} = -\frac{1}{13} \). 
The slope of the tangent line at any point on the curve is given by the derivative of the curve \( y = 5x^2 + \alpha x + \beta \): \[ \frac{dy}{dx} = 10x + \alpha. \] At the point \( (1, 3) \), the slope of the tangent line is: \[ \frac{dy}{dx} = 10(1) + \alpha = 10 + \alpha. \] Since the normal line and the tangent line are perpendicular, the product of their slopes must be \( -1 \): \[ m_{\text{normal}} \times m_{\text{tangent}} = -1. \] Substitute \( m_{\text{normal}} = -\frac{1}{13} \) and \( m_{\text{tangent}} = 10 + \alpha \): \[ \left( -\frac{1}{13} \right) \times (10 + \alpha) = -1. \] Simplifying this equation: \[ \frac{1}{13} \times (10 + \alpha) = 1 \quad \Rightarrow \quad 10 + \alpha = 13 \quad \Rightarrow \quad \alpha = 3. \]
Now substitute \( x = 1 \) and \( y = 3 \) into the equation of the curve to find \( \beta \): \[ 3 = 5(1)^2 + 3(1) + \beta \quad \Rightarrow \quad 3 = 5 + 3 + \beta \quad \Rightarrow \quad \beta = -5. \] Thus, \( \alpha = 3 \) and \( \beta = -5 \), so: \[ \alpha \beta = 3 \times (-5) = -15. \]

The correct option is (E) : \(-15\)

Answer 2

We are given the equation of the normal line to the curve \(y = 5x^2 + ax + \beta\) at the point (1, 3) as \(x + 13y = 40\). Our goal is to find the value of \(\alpha\beta\).

Since the point (1, 3) lies on the curve, we can substitute \(x = 1\) and \(y = 3\) into the equation of the curve: \[3 = 5(1)^2 + a(1) + \beta\] \[3 = 5 + a + \beta\] \[a + \beta = -2 \quad (*)\]

Now, let's find the derivative of \(y\) with respect to \(x\): \[\frac{dy}{dx} = \frac{d}{dx}(5x^2 + ax + \beta) = 10x + a\]

The slope of the tangent line at the point (1, 3) is: \[m_t = \frac{dy}{dx}\Big|_{x=1} = 10(1) + a = 10 + a\]

The equation of the normal line is \(x + 13y = 40\). We can rewrite this in slope-intercept form to find the slope of the normal line: \[13y = -x + 40\] \[y = -\frac{1}{13}x + \frac{40}{13}\] Thus, the slope of the normal line is \(m_n = -\frac{1}{13}\).

Since the normal line is perpendicular to the tangent line, the product of their slopes is -1: \[m_t \cdot m_n = -1\] \[(10 + a)\left(-\frac{1}{13}\right) = -1\] \[10 + a = 13\] \[a = 3\]

Now, substitute \(a = 3\) into equation \((*)\): \[3 + \beta = -2\] \[\beta = -5\]

Finally, we find the value of \(\alpha\beta = a\beta = (3)(-5) = -15\).

Therefore, the value of \(a\beta\) is -15.