Answer :

Curve: y = 4x^{3} – 3x +5

Let the required normal be at (x_{1}, y_{1}) on the curve

Slope of tangent of the curve

∴Slope of normal at any point (x_{1}, y_{1})

Slope of given line = 9

As, given line is perpendicular to the normal, therefore,

Now, put x_{1} = +1 and -1 on the equation of curve to find y_{1}

Equation of curve: y = 4x^{3} – 3x +5

Put x = 1, we get,

y = 4(1)^{3} – 3(1) + 5

⇒ y = 4 – 3 + 5

⇒ y = 6

So, point is (1, 6)

Now put x = -1 in equation of curve we get,

y = 4(-1)^{3} – 3(-1) + 5

⇒ y = -4 + 3 + 5

⇒ y = 4

So, point is (-1, 4)

So, we get the points as (1, 6) and (-1, 4)

Therefore, equations of normal are:

⇒9y – 54 = –x + 1 and 9y – 36 = –x – 1

⇒x + 9y = 55 and x + 9y = 35

So, required equations are x + 9y = 55 and x + 9y = 35.

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