A normal line is drawn to the curve y = \frac{x^2 - 5}{x - 3} at the point where x = 2. Wh… — JAMB UTME

A normal line is drawn to the curve $y = \frac{x^2 - 5}{x - 3}$ at the point where $x = 2$ . What is the $x$ -intercept of this normal line?

Answer: -1

The

x

-intercept of the normal line is

-1

Substituting

x = 2

into the curve yields the point

(2, 1)

. Differentiation via the quotient rule gives

\frac{dy}{dx} = \frac{x^2 - 6x + 5}{(x - 3)^2}

. At

x = 2

, the tangent slope is

-3

, making the normal slope

\frac{1}{3}

. The line equation

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

simplifies to

x - 3y + 1 = 0

. Setting

y = 0

gives

x = -1

Calculate the y-coordinate of the point of tangency

y = 1

Substitute

x = 2

into the curve equation

y = \frac{x^2 - 5}{x - 3}

to get the point

(2, 1)

Differentiate the function using the quotient rule

\frac{dy}{dx} = \frac{x^2 - 6x + 5}{(x - 3)^2}

Applying

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

u = x^2 - 5

and

v = x - 3

Find the tangent gradient at

x = 2

m_t = -3

Evaluate

\frac{dy}{dx}

x = 2

to obtain

m_t = \frac{2^2 - 6(2) + 5}{(2-3)^2} = -3

Calculate the gradient of the normal line

m_n = \frac{1}{3}

The normal line is perpendicular to the tangent line, so

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

Derive the line equation for the normal

x - 3y + 1 = 0

Use point-slope form

y - y_1 = m_n(x - x_1)

with

(2, 1)

and

m_n = \frac{1}{3}

Find the x-intercept

x = -1

Set

y = 0

in the normal equation

x - 3(0) + 1 = 0

, giving

x = -1

Equation of normal line to a rational curve and finding its axis intercepts

Question