What are the coordinates of the point on the curve y = 2x^2 - 5x + 1 where the tangent lin… — JAMB UTME

What are the coordinates of the point on the curve $y = 2x^2 - 5x + 1$ where the tangent line is perpendicular to the line $x + 3y - 4 = 0$ ?

The point on the curve is

(2, -1)

Rearranging the line equation

x + 3y - 4 = 0

gives a gradient of

-\frac{1}{3}

. The tangent line is perpendicular, so its gradient must be

3

. Differentiating

y = 2x^2 - 5x + 1

gives

\frac{dy}{dx} = 4x - 5

. Setting

4x - 5 = 3

gives

x = 2

. Substituting

x = 2

into the curve equation yields

y = 2(2)^2 - 5(2) + 1 = -1

. Thus, the point is

(2, -1)

Determine the gradient of the given line.

Rearranging

x + 3y - 4 = 0

into slope-intercept form gives

y = -\frac{1}{3}x + \frac{4}{3}

, so the gradient is

m_1 = -\frac{1}{3}

The slope of a linear equation

Ax + By + C = 0

-\frac{A}{B}

Calculate the gradient of the tangent line.

Since the tangent line is perpendicular to the given line, its gradient is

m_T = -\frac{1}{m_1} = 3

Perpendicular lines have gradients whose product is

-1

(

m_1 \cdot m_2 = -1

Find the derivative of the curve and set it equal to the tangent gradient.

\frac{dy}{dx} = \frac{d}{dx}(2x^2 - 5x + 1) = 4x - 5

. Setting

4x - 5 = 3

yields

4x = 8 \implies x = 2

The derivative

\frac{dy}{dx}

gives the gradient of the tangent to the curve at any point

x

Substitute the

x

-coordinate into the original curve equation to find

y

y = 2(2)^2 - 5(2) + 1 = 8 - 10 + 1 = -1

The point lies on the curve, so its coordinates must satisfy the curve's equation.

Tangents and Normals to Curves

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