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Answer 1

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Answer 2

Answer :-

Given

♦ Curve → y = x³ + 2x + 6

♦ Equation of line → x + 14y +4 = 0

According to the question :-

We have to find the normal of curve y = x³ + 2x + 6 , which is parallel to the line x + 14y + 4 .

Step 1 :- Finding out the slope ( as parallel lines have same slope)

in a equation

y = mx + c

m = slope of the line

Here equation of line

= x + 14y + 4

By converting it

→ 14y = -x - 4

$\longrightarrow y = \dfrac{-1}{14}x -\dfrac{4}{14}$

So slope of line = $\dfrac{-1}{14}$

Step 2 :- Differentiation of curve

y = x³ + 2x + 6

$\implies \dfrac{dy}{dx}(x^3 + 2x + 6)$

$\implies \dfrac{dy}{dx}(x^3) + \dfrac{dy}{dx}(2x) + \dfrac{dy}{dx}(6)$

$\implies (3)x^{3-1} + (1)2x^{1-1} + 0$

$\implies \dfrac{dy}{dx}(x^3 + 2x + 6) = 3x^2 + 2$

Step 3 :- Now by using Formula for slope of Normal .

$Slope\: of\: Normal = -\dfrac{1}{\left(\dfrac{dy}{dx}\right)}$

$\implies \dfrac{-1}{14} = -\dfrac{1}{3x^2 + 2}$

$\implies 3x^2 + 2 = 14$

$\implies 3x^2 = 12$

$\implies x^2 = 4$

$\implies x = \pm 2$

Step 4 :- Replacing value of x in y = x³ + 2x + 6

When x = 2

→ y = 2³ + 2(2) + 6

→ y = 8 + 4 + 6

→ y = 18

When x = -2

→ y = (-2)³ + 2(-2) + 6

→ y = -8 - 4 + 6

→ y = -6

Step 5 :- Finding out the equation of line

Normal by Point ( 2 , 18)

$\implies y - 18 = \dfrac{-1}{14} (x-2)$

$\implies 14y - 252 = - x + 2$

$\implies x + 14y -254 = 0$

Normal by Point ( -2 , -6 )

$\implies y - (- 6) = \dfrac{-1}{14} (x -(-2))$

$\implies y + 6 = \dfrac{-1}{14} (x + 2)$

$\implies 14y + 84 = -x - 2$

$\implies x + 14y + 86 = 0$