Question

Find the equation of the normals to the curve y = x ^3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

Answer 1

HELLO DEAR,

The given line is x + 14y + 4 = 0.
y = -x/14 - 2/7

therefore,slop of the given line = -1/14

so,
the slope of the required normal = -1/14-----( 1 )
let the pojntof contact be (h , k).

Now, y = x³ + 2x + 6
differentiating y w.r.t. x,
dy/dx = 3x² + 2

$\bold{\frac{dy}{dx}_{h,k} = 3h^2+ 2}$.

therefore, slope of normal = $\bold{\frac{-1}{\frac{dy}{dx}_{h,k}} = \frac{-1}{3h^2 + 2}}$.---( 2 )

from------( 1 ) & ------( 2 )
$\bold{\frac{-1}{3h^2+2} = \frac{-1}{14}}$

$\bold{3h^2 + 2 = 14}$

$\bold{h^2 = \frac{12}{3}}$

$\bold{h = ±2}$

put h = 2 ,we get k = 8 + 4 + 6 = 18
put h = -2 , we get k = -8 -4 + 6 = -6

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

so, the equation of the required normal at (2 , 18)is, $\bold{\frac{y - 18}{x - 2} = \frac{-1}{14}}$,
i.e. 14(y - 18) = (2 - x)
14y + x = 254,

the, equation of the required normal at ( -2 , -6) is,$\bold{\frac{y + 6}{x + 2} = \frac{-1}{14}}$
i.e., 14(y + 6) = -x - 2
14y + x = -86

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

Step-by-step explanation:

Hope you understood.