At what point on the curve y = 2x^2 - 8x + 5 is the normal line parallel to the straight l… — JAMB UTME

At what point on the curve $y = 2x^2 - 8x + 5$ is the normal line parallel to the straight line $x + 4y - 7 = 0$ ?

The point of contact on the curve is

(3, -1)

The line

x + 4y - 7 = 0

has a slope of

-\frac{1}{4}

. Since the normal line is parallel to this line, the normal gradient is

m_n = -\frac{1}{4}

. Consequently, the tangent gradient must be

m_t = 4

because

m_t \cdot m_n = -1

. Equating the derivative

\frac{dy}{dx} = 4x - 8

4

yields

x = 3

. Substituting

x = 3

into the curve equation

y = 2x^2 - 8x + 5

gives

y = -1

, yielding the point

(3, -1)

Find the gradient of the given straight line.

Rearranging

x + 4y - 7 = 0

into slope-intercept form gives

y = -\frac{1}{4}x + \frac{7}{4}

, so the line's gradient is

m = -\frac{1}{4}

Parallel lines have equal gradients, so the normal line to the curve must have gradient

m_n = -\frac{1}{4}

Determine the required gradient of the tangent line.

Since

m_t \cdot m_n = -1

, we have

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

The tangent and normal lines are perpendicular to each other.

Differentiate the curve's equation to find the

x

-coordinate.

\frac{dy}{dx} = 4x - 8

. Setting

\frac{dy}{dx} = 4

gives

4x - 8 = 4 \implies 4x = 12 \implies x = 3

The derivative represents the slope of the tangent line at any point

x

Substitute

x = 3

back into the curve's equation to find

y

y = 2(3)^2 - 8(3) + 5 = 2(9) - 24 + 5 = 18 - 24 + 5 = -1

The point of contact lies on the original curve.

Relationship between gradients of parallel lines, tangent lines, and normal lines to a curve

Question