Question 1151870
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The graph of {{{y^2=2x}}} is symmetrical with respect to the x-axis.  So we can find the equations of the tangent lines for the portion of the graph with y positive; then any tangent line with equation {{{y=mx+b}}} will have a corresponding tangent line with equation {{{y=-(mx+b)}}}, or {{{y=-mx-b}}}.<br>
The equation for the portion of the graph with y positive is<br>
{{{y = sqrt(2x) = (2x)^(1/2)}}}<br>
Take the derivative, using the power and chain rules, to find the slope:<br>
{{{dy/dx =(1/2)(2x)^(-1/2)(2) = (2x)^(-1/2)}}}<br>
For each value of x, the tangent line will have slope {{{(2x)^(-1/2)}}}; and the y value will be {{{(2x)^(1/2)}}}<br>
For a given value of x, find the expression for the y-intercept b for the tangent line with {{{y = (2x)^(1/2)}}} and slope {{{(2x)^(-1/2)}}}.<br>
{{{y = mx+b}}}
{{{(2x)^(1/2) = ((2x)^(-1/2))x + b}}}
{{{sqrt(2)*sqrt(x) = (1/(sqrt(2)*sqrt(x)))*x + b}}}
{{{b = sqrt(2)*sqrt(x) - (1/(sqrt(2)*sqrt(x)))*x}}}
{{{b = sqrt(2)*sqrt(x) - (sqrt(x)/sqrt(2))}}}
{{{b = sqrt(x)(sqrt(2)-1/sqrt(2))}}}
{{{b = sqrt(x)(2/sqrt(2)-1/sqrt(2))}}}
{{{b = sqrt(x)(1/sqrt(2))}}}
{{{b = sqrt(x/2)}}}<br>
So for a given value of x, the tangent line in the first quadrant has slope {{{1/sqrt(2x)}}} and y-intercept {{{sqrt(x/2)}}}.<br>
The equation is then<br>
{{{y = (1/sqrt(2x))x + sqrt(x/2)}}}<br>
Choosing some "nice" values of x that make {{{sqrt(2x)}}} and {{{sqrt(x/2)}}} rational....<br>
x = 1/2:  {{{y = x+1/2}}}
x = 2:  {{{y = (1/2)x+1}}}
x = 9/2:  {{{y = (1/3)x+3/2}}}
x = 8:  {{{y = (1/4)+2}}}<br>
And then there are corresponding tangent lines on the other branch of the parabola:<br>
x = 1/2:  {{{y = -x-1/2}}}
x = 2:  {{{y = (-1/2)x-1}}}
x = 9/2:  {{{y = (-1/3)x-3/2}}}
x = 8:  {{{y = (-1/4)-2}}}<br>
A graph of the parabola and the tangent lines for x=1/2 -- x+1/2 and -x-1/2:<br>
{{{graph(800,400,-2,10,-5,5,sqrt(2x),-sqrt(2x),x+1/2,-x-1/2)}}}
A graph of the parabola and the tangent lines for x=2 -- (1/2)x+1 and (-1/2)x-1:<br>
{{{graph(800,400,-2,10,-5,5,sqrt(2x),-sqrt(2x),(1/2)x+1,(-1/2)x-1)}}}
A graph of the parabola and the tangent lines for x=8 -- (1/4)x+2 and (-1/4)x-2:<br>
{{{graph(800,400,-2,10,-5,5,sqrt(2x),-sqrt(2x),(1/4)x+2,(-1/4)x-2)}}}