What normal to the curve y = x^2 forms the shortest chord

Dear student


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First, note that the derivative of y=x^2 is y'=2x, which means that the slope of the tangent at (x,x^2) is 2x so the slope of the normal at (x,x^2) is -1/2x. 

An equation of the normal at (a,a^2) is 


Observe that the line intersection y=x^2 at the solutions of the system 
{y=x^2,y-a^2=-1/(2a) (x-a)}: 

(a,a^2) and (-1/(2a)-a,1+1/(4a^2)+a^2). 

We are trying to find a value of a that minimizes the distance between these two points. That is, we want to minimize 


Differentiating gives us 

=(2a^1-1)(4a^2+1)^2 / (4a^5) 

and solving 

l'(a)=0 gives us a=-1/\sqrt{2} or a=1/\sqrt{2}. 
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