a) Find the equation of the normal to the parabola $y=x^3$, where $x =4$.b) Find the coordinates of the point at which this normal cuts the parabola again.

c) Show that the distance between the two points where the parabola cuts the normal is 1$6*\sqrt{2}$ units.

Answer

The slope of the tangent line at point $x=4$ to $y=x^2/8$ is

$y'(x) = 2x/8 =x/4$, which at $x=4$ gives $y'(4) =1$

If m is the slope of the tangent and m1 is the slope of the normal we have $m*m1=-1$.

$m1 =-1/1 =-1$.

Now at $x = 4$, $y = 4*4/8 =2$. The equation of line through (4,2) with a slope $m1=-1$ is

$y -2 =-1(x-4)$

$y = -x +4 +2$

$y =- x +6$

Now for the second intersection point we have

$x^2/8 = -x +6$

$x^2 +8x -48 =0$

$x1 = -4+ \sqrt{(16+48)} =-4+8 =4$

$x2 =-1-\sqrt{(16 +48)} =-4 -8 =-12$

$y2 = (-12)^2/8 = 18$

Therefore the second intersection is at (-12, 18)

The distance between these two points is

$d =\sqrt{((x1-x2)^2 +(y2-y1)^2)} =\sqrt{(4+12)^2 +(2-18)^2)} =$

$=\sqrt{(256 +256)}= 16*\sqrt{(2)}$