What is the equation of the normal to the curve y = 2x^2 - 3x + 1 at the point where x = 2… — JAMB UTME

What is the equation of the normal to the curve $y = 2x^2 - 3x + 1$ at the point where $x = 2$ ?

x + 5y - 17 = 0

Evaluating

y = 2x^2 - 3x + 1

x = 2

gives the point

(2, 3)

. Differentiating gives

\frac{dy}{dx} = 4x - 3

, which evaluates to

5

x = 2

. The normal line gradient is therefore

-\frac{1}{5}

. Using the point-slope formula

y - 3 = -\frac{1}{5}(x - 2)

and rearranging to standard linear form results in

x + 5y - 17 = 0

Find the

y

-coordinate of the point of contact on the curve when

x = 2

Substitute

x = 2

into

y = 2x^2 - 3x + 1

y = 2(2)^2 - 3(2) + 1 = 8 - 6 + 1 = 3

. The point is

(2, 3)

The line of normal passes through the point of tangency on the curve.

Differentiate

y = 2x^2 - 3x + 1

with respect to

x

to find the gradient of the tangent.

\frac{dy}{dx} = 4x - 3

. At

x = 2

m_t = 4(2) - 3 = 5

The derivative of a curve evaluated at a point yields the gradient of the tangent line at that point.

Determine the gradient of the normal line.

Since the normal is perpendicular to the tangent,

m_n = -\frac{1}{m_t} = -\frac{1}{5}

Perpendicular lines have gradients whose product is

-1

(

m_t \cdot m_n = -1

Form the equation of the normal line passing through

(2, 3)

with gradient

m_n = -\frac{1}{5}

y - 3 = -\frac{1}{5}(x - 2) \implies 5(y - 3) = -(x - 2) \implies 5y - 15 = -x + 2 \implies x + 5y - 17 = 0

Applying the point-slope form

y - y_1 = m(x - x_1)

establishes the linear equation.

Tangents and Normals to Curves

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