Question

Equation of the normal to the curve \( y = x^2 + x \) at the point \( (1,2) \) is:

• A. \( x - 3y + 5 = 0 \)
• B. \( x + 3y + 7 = 0 \)
• C. \( x + 3y + 5 = 0 \)
• D. \( x + 3y - 7 = 0 \)

Hint:

For normal equations, first find the derivative to get the tangent slope, then use its negative reciprocal to find the normal's equation. Always substitute the given point correctly.

Answer 1

The Correct Option is D

Step 1: Differentiate the given function The given curve equation is: \[ y = x^2 + x. \] Differentiating both sides with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx} (x^2 + x) = 2x + 1. \] Thus, the slope of the tangent at any point \( x \) is: \[ m_t = 2x + 1. \]
Step 2: Find the slope at \( (1,2) \) Substituting \( x = 1 \): \[ m_t = 2(1) + 1 = 3. \] Since the slope of the normal is the negative reciprocal of the tangent slope, the normal’s slope \( m_n \) is: \[ m_n = -\frac{1}{3}. \]
Step 3: Equation of the normal line The equation of a line with slope \( m \) passing through a point \( (x_1, y_1) \) is given by: \[ y - y_1 = m (x - x_1). \] Substituting \( (x_1, y_1) = (1,2) \) and \( m_n = -\frac{1}{3} \): \[ y - 2 = -\frac{1}{3} (x - 1). \] Multiplying both sides by 3 to eliminate the fraction: \[ 3(y - 2) = -(x - 1). \] \[ 3y - 6 = -x + 1. \] \[ x + 3y - 7 = 0. \] Thus, the correct answer is option (4) \( x + 3y - 7 = 0 \).

Answer 2

To find the equation of the normal to the curve \( y = x^2 + x \) at the point \( (1,2) \), we must follow these steps:

Step 1: Determine the derivative to find the slope of the tangent at \( x = 1 \).
The derivative \( \frac{dy}{dx} \) of \( y = x^2 + x \) is:

$\frac{d}{y}=\frac{d}{x}(x^2+x)=2x+1$

Step 2: Calculate the slope of the tangent at \( x = 1 \).

$m={y\text{'}}_{|}x=1=2\left(1\right)+1=3$

Step 3: Determine the slope of the normal line.
The slope of the normal line is the negative reciprocal of the slope of the tangent:

$m=-\frac{1}{3}$

Step 4: Use the point-slope form to write the equation of the normal line through the point \( (1,2) \).
The point-slope form of a line is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point on the line.

$y-2=-\frac{1}{3}(x-1)$

Simplify to obtain:

$3(y-2)=-(x-1)$

$3y-6=-x+1$

$x+3y-7=0$

Thus, the equation of the normal is \( x + 3y - 7 = 0 \).