A normal line is drawn to the curve y = \frac{x^2 + 3}{x - 1} at the point where x = 2. Ca… — JAMB UTME

A normal line is drawn to the curve $y = \frac{x^2 + 3}{x - 1}$ at the point where $x = 2$ . Calculate the $x$ -intercept of this normal line.

Answer: -19

Answer

The x-intercept of the normal line is -19.

Substituting

x = 2

into the curve function gives

y = 7

. Differentiating via the quotient rule yields

\frac{dy}{dx} = \frac{x^2 - 2x - 3}{(x - 1)^2}

, which equals

-3

x = 2

. The perpendicular slope of the normal line is

\frac{1}{3}

. Constructing the line equation through

(2, 7)

gives

x - 3y + 19 = 0

. Setting

y = 0

produces

x = -19

Step-by-Step Solution

Find the point of contact by substituting

x = 2

into the curve equation

y = \frac{2^2 + 3}{2 - 1} = \frac{7}{1} = 7

, so the point is

(2, 7)

The line is drawn at

x = 2

, so we need the full coordinate pair

(x_1, y_1)

Differentiate the curve

y = \frac{x^2 + 3}{x - 1}

using the quotient rule

\frac{dy}{dx} = \frac{(x - 1)(2x) - (x^2 + 3)(1)}{(x - 1)^2} = \frac{x^2 - 2x - 3}{(x - 1)^2}

The derivative gives the gradient function of the curve

Evaluate the tangent slope

m_t

x = 2

m_t = \frac{2^2 - 2(2) - 3}{(2 - 1)^2} = \frac{-3}{1} = -3

Evaluating the derivative yields the gradient of the tangent at the point

Determine the slope of the normal line

m_n

m_n = -\frac{1}{m_t} = -\frac{1}{-3} = \frac{1}{3}

Normal lines are perpendicular to tangent lines, so

m_n \cdot m_t = -1

Formulate the equation of the normal line passing through

(2, 7)

y - 7 = \frac{1}{3}(x - 2) \implies 3y - 21 = x - 2 \implies x - 3y + 19 = 0

Use the point-slope form

y - y_1 = m(x - x_1)

Find the

x

-intercept by setting

y = 0

x - 3(0) + 19 = 0 \implies x = -19

The

x

-intercept occurs where the line crosses the

x

-axis (

y = 0

)

Key Concept

Finding the equation and x-intercept of a normal line to a curve

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