Question

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

Answer 1

Consider ( h, k ) be the point on curve from tangent.

The given equation of the curve is,

y= x 3 +2x+6

The point ( h, k ) is on the curve, so the point satisfies the equation of curve.

k= h 3 +2h+6(1)

Slope of tangent at ( h, k ) is,

dy dx | ( h, k ) =3 h 2 +2

Now, the formula for the slope of the normal at ( h, k ) is,

−1 ( dy dx ) | ( h, k ) = −1 3 h 2 +2

Given, the normal is parallel to the equation of the line x+14y+4=0. So, slope of normal is equal to slope of the line.

x+14y+4=0 y=( −1 14 )x−( 4 14 )

The above equation is of the form of y=mx+c where, m is the slope of line. So, slope of line is ( −1 14 ).

Now, slope of normal at ( h, k ) is equal to the slope of line x+14y+4=0.

−1 3 h 2 +2 = −1 14 14=3 h 2 +2 h 2 = 12 3 h=±2

Now, substitute h=2 in the equation (1),

k= h 3 +2h+6 k= 2 3 +2( 2 )+6 =18

Here, the point is ( 2, 18 ).

Now, substitute h=−2 in the equation (1),

k= h 3 +2h+6 k= ( −2 ) 3 +2( −2 )+6 =−6

Here, the point is ( −2, −6 ).

The equation of the line at ( x 1 ,  y 1 ) and having slope m is ( y− y 1 )=m( x− x 1 ).

So that equation of the normal through ( 2, 18 ) and having slope −1 14 is,

y−18= −1 14 ( x−2 ) x+14y−254=0

Now the equation of the normal through ( −2, −6 ) and having slope −1 14 is,

y−( −6 )= −1 14 ( x−( −2 ) ) x+14y+86=0

Thus, the required equations of normals are x+14y−254=0 and x+14y+86=0.