Question 1039398
The slope of the tangent lines is equal to the value of the derivative. 
So first find the derivative of each,
{{{x^2/2+y^2=r}}}
Implicit differentiation,
{{{(1/2)(2xdx)+2ydy=0}}}
{{{2ydy=-xdx}}}
{{{(dy/dx)[1]=-x/(2y)}}}
and
{{{y=cx^2}}}
{{{(dy/dx)[2]=2cx}}}
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So then assume that the point intersect at ({{{x[0]}}},{{{y[0]}}}).
Then,
{{{x[0]^2/2+y[0]^2=r}}}
and
{{{y[0]=c*x[0]^2}}}
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The first tangent line is,
{{{y-y[0]=-(x[0]/(2y[0]))(x-x[0])}}}}
The second tangent line is,
{{{y-y[0]=2cx[0](x-x[0])}}}
In order for the tangent lines to be perpendicular, the slopes must be negative reciprocals,
{{{m[1]*m[2]=-1}}}
{{{-(x[0]/(2y[0]))(2cx[0])=-1}}}
{{{(cx[0]^2)/(y[0])=1}}}
Since {{{y[0]=cx[0]^2}}},
{{{y[0]/y[0]=1}}}
{{{1=1}}}
So they are perpendicular to each other.
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*[illustration dsa1.JPG].
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Differentiate,
{{{v=dx/dt=1/(2t+1)-2t}}}
{{{a=dv/dt=-4/(2t+1)^2-2}}}
Let's graph the derivatives to see,
*[illustration dsa2.JPG].
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As you can see the velocity is positive for a bit and then becomes negative.
Let's find when it equals zero.
{{{1/(2t+1)-2t=0}}}
{{{1/(2t+1)=2t}}}
{{{2t(2t+1)=1}}}
{{{4t^2+wt-1=0}}}
{{{4(t^2+t/2)-1=0}}}
{{{4(t^2+t/2+1/16)-1-4(1/16)=0}}}
{{{4(t+1/4)^2=5/4}}}
{{{2(t+1/4)=sqrt(5)/2}}}
{{{t+1/4=sqrt(5)/4}}}
{{{t=(sqrt(5)-1)/4}}}
So velocity is positive or zero for {{{0<=t<(sqrt(5)-1)/4}}}
and negative for {{{t>(sqrt(5)-1)/4}}}
The acceleration is always negative so the velocity steadily increases in the negative direction.