Tangents and Normals

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.

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
$d =\sqrt{((x1-x2)^2 +(y2-y1)^2)} =\sqrt{(4+12)^2 +(2-18)^2)} =$
$=\sqrt{(256 +256)}= 16*\sqrt{(2)}$


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