What is the y-intercept of the normal line to the curve y = \frac{2x - 1}{x + 1} at the po… — JAMB UTME

What is the $y$ -intercept of the normal line to the curve $y = \frac{2x - 1}{x + 1}$ at the point where the curve crosses the $x$ -axis?

$\frac{3}{8}$ Answer
B
$-\frac{2}{3}$
C
$\frac{9}{8}$
D
$\frac{2}{3}$

Answer

The

y

-intercept of the normal line is

\frac{3}{8}

To find the

y

-intercept of the normal line, set

y = 0

to find the point of contact on the

x

-axis, which gives

\left(\frac{1}{2}, 0\right)

. Differentiating

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

gives

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

. At

x = \frac{1}{2}

, the tangent slope is

\frac{4}{3}

, making the normal slope

-\frac{3}{4}

. The equation of the normal line is

y = -\frac{3}{4}x + \frac{3}{8}

, so its

y

-intercept is

\frac{3}{8}

Step-by-Step Solution

Find the point of intersection of the curve with the

x

-axis

Set

y = 0 \implies \frac{2x - 1}{x + 1} = 0 \implies 2x - 1 = 0 \implies x = \frac{1}{2}

. The point is

\left(\frac{1}{2}, 0\right)

The normal line is drawn at the point where the curve crosses the

x

-axis.

Differentiate

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

using the quotient rule

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

The derivative provides the gradient function of the tangent line to the curve.

Evaluate the gradient of the tangent and normal lines at

x = \frac{1}{2}

Tangent gradient

m_t = \frac{3}{\left(\frac{1}{2} + 1\right)^2} = \frac{3}{\frac{9}{4}} = \frac{4}{3}

. Normal gradient

m_n = -\frac{1}{m_t} = -\frac{3}{4}

The normal line is perpendicular to the tangent line at the point of contact.

Determine the equation of the normal line and evaluate its

y

-intercept

Using

y - 0 = -\frac{3}{4}\left(x - \frac{1}{2}\right) \implies y = -\frac{3}{4}x + \frac{3}{8}

. Setting

x = 0

gives

y = \frac{3}{8}

The

y

-intercept is the value of

y

when

x = 0

on the line.

Key Concept

Tangents and Normals to Curves

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