Question 578107
prove that the line y+x=-2 is a tangent to the curve y^2 = 8x.
<pre>
A line which intersects a parabola exactly once and which is not 
parallel to the axis of symmetry, is tangent to the parabola.  


{{{drawing(300,300,-10,10,-10,10,

graph(300,300,-10,10,-10,10,-2-x),

graph(300,300,-10,10,-10,10,sqrt(8x)),

graph(300,300,-10,10,-10,10,-sqrt(8x)) )}}}

y + x = -2 is a tangent to the curve y² = 8x

The axis of symmetry of that parabola is the 
x-axis, and the line y + x = -2 is not parallel
to the x-axis, so if it intersects the parabola
exactly once, then it is tangent to the parabola.

We solve the equation of the line for y

    y + x = -2 - x
        y = -2 - x

And we substitute (-2 - x) for y in

            y² = 8x
     (-2 - x)² = 8x
   4 + 4x + x² = 8x 

   x² - 4x + 4 = 0

(x - 2)(x - 2) = 0

x - 2 = 0,  x - 2 = 0
    x = 2       x = 2

Th fact that -2 is a double root for x shows that 
there is just one point of intersection and so the
line is tangent to the parabola at the point were
x = 2, which has y-coordinate y = -2 - (2) = -4

or the point (2,-4) is the point of tangency.

{{{drawing(300,300,-10,10,-10,10, locate(2,-2.7,"(2,-4)"),

graph(300,300,-10,10,-10,10,-2-x),

graph(300,300,-10,10,-10,10,sqrt(8x)),

graph(300,300,-10,10,-10,10,-sqrt(8x)) )}}}



Edwin</pre>