58. Find the equation of the normals to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

The given eqn of the curve is y=x3+2x+6 _____(1) slope of tangent to the curve, dydx=3x2+2 so, slope of normal to the curve =−13x2+2 Now, the line x+14y+4=0→y=−114x−414 compared to y=mx+c gives slope of line = −114 As the normal is parallel to the line ⇒ −13x2+2=−114 →3x2+2=14 ⇒3x2=12 →x2=4 ⇒x=±2 When x=2,y=(2)3+2(2)+6=8+4+6=18 and when x=−2, y=(−2)3+2(−2)+6=−8−4+6=−6 The point of contact of the normal are (2, 18) and (-2, -6) Hence the eqn of normal are y−18=−114(x−2) and y−(−6)=−114[x−(−2)] ⇒14y−252=−x+2 ⇒y+6=−114(x+2) ⇒x+14y−254=0 ⇒14y+84=−x−2 ⇒x+14y+86=0.

58. Find the equation of the normals to the curve y = x³ + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

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Vishal Baghel | Contributor-Level 10

9 months ago

The given eqⁿ of the curve is $y = x^{3} + 2 x + 6$ _____(1)

slope of tangent to the curve, $\frac{d y}{d x} = 3 x^{2} + 2$

so, slope of normal to the curve $= \frac{- 1}{3 x^{2} + 2}$

Now, the line $x + 14 y + 4 = 0 \to y = - \frac{1}{14} x - \frac{4}{14}$ compared to $y = m x + c$ gives

slope of line = $- \frac{1}{14}$

As the normal is parallel to the line

$\Rightarrow \frac{- 1}{3 x^{2} + 2} = - \frac{1}{14}$

$\to 3 x^{2} + 2 = 14$

$\Rightarrow 3 x^{2} = 12$

$\to x^{2} = 4$

$\Rightarrow x = \pm 2$

When $x = 2, y = {(2)}^{3} + 2 (2) + 6 = 8 + 4 + 6 = 18$

and when $x = - 2, y = {(- 2)}^{3} + 2 (- 2) + 6 = - 8 - 4 + 6 = - 6$

The point of contact of the normal are (2, 18) and (-2, -6)

Hence the eqⁿ of normal are

$y - 18 = - \frac{1}{14} (x - 2)$ and $y - (- 6) = \frac{- 1}{14} [x - (- 2)]$

$\Rightarrow 14 y - 252 = - x + 2$ $\Rightarrow y + 6 = \frac{- 1}{14} (x + 2)$

$\Rightarrow x + 14 y - 254 = 0$ $\Rightarrow 14 y + 84 = - x - 2$

$\Rightarrow x + 14 y + 86 = 0 .$

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