Question

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

Answer 1

The equation of the given curve is $y=x^3+2x+6$

The slope fo the tangent to the given curve at any point (x,y) is given by

$\frac{dy}{dx}=3x^2+2$

$\therefore$ Slope of the normal to the given curve at any point

$(x,y)=\frac{-1}{Slopeofthetangentatthepoint(x,y)}=\frac{-1}{3x^2+2}$

The equation of the given line x+14y+4=0.

$\Rightarrow y=\frac{-x}{14}-\frac{4}{14}$(which is of the form y=mx+c)

$\therefore$ Slope of the given line $=\frac{-1}{14}$

If the normal is parallel to the line x+14y+4=0, then we must have the slope of the normal being equal to the line.

$\therefore \frac{-1}{3x^2+2}=-\frac{1}{14}\Rightarrow 3x^2+2=14\Rightarrow 3x^2=12\Rightarrow x^2=4\Rightarrow x=\pm 2$

when x=2, y=8+4+6=18. When x=-2, y=-8-4+6=-6

Therefore, there are two normals to the given curve with slope $\frac{-1}{14}$ and

passing through the points (2,18) and (-2,-6).

Thus, the equation of the normal passing through (2,18) is given by

$y-18=\frac{-1}{14}(x-2)\Rightarrow 14y-252=-x+2\Rightarrow x+14y-254=0$

and the equation of the normal passing through (-2,-6) is given by

$y-(-6)=\frac{-1}{14}[x-(-2\left)\right]\Rightarrow y+6=\frac{-1}{14}(x+2)$

$\Rightarrow 14y+84=-x-2\Rightarrow x+14y+86=0$

Hence, the equation of the normals to the given curve, which are parallel to the given line are x+14y-254=0 and x+14y+86=0.