A Fairly Normal Asymmetry Problem

I made a similar problem Can you minimize it there you had to minimze the length of the chord. So I will post a similar solution here.

We take a point ${ P }_{ 1 }=({ x }_{ 0 },{ x }_{ 0 }^{ 2 })$ on the parabola. Then slope of tangent is $=2{ x }_{ 0 }$

Hence slope of normal is $\frac { -1 }{ 2{ x }_{ 0 } }$

So equation of normal is :

$(x-{ x }_{ 0 })=-2{ x }_{ 0 }(y-{ x }_{ 0 }^{ 2 })$

Solving it with the parabola we get :

$x={ x }_{ 0 },-\frac { 1 }{ 2{ x }_{ 0 } } -{ x }_{ 0 }$

So the other point is :

${ P }_{ 2 }=(-\frac { 1 }{ 2{ x }_{ 0 } } -{ x }_{ 0 },{ (-\frac { 1 }{ 2{ x }_{ 0 } } -{ x }_{ 0 } })^{ 2 }$

Hence the area bounded by ${P}_{1}{P}_{2}$ and the parabola is :

$\int _{ (-\frac { 1 }{ 2{ x }_{ 0 } } +{ x }_{ 0 }) }^{ { x }_{ 0 } }{ { (x }_{ 0 }^{ 2 }+\frac { 1 }{ 2 } -x-{ x }^{ 2 })dx }$ (using the equation of normal)

Evaluating it we get :

$A=\frac { 4 }{ 3 } { ({ x }_{ 0 }+\frac { 1 }{ 4{ x }_{ 0 } } ) }^{ 3 }$

By Applying AM-GM inequality we get :

$\frac { { x }_{ 0 }+\frac { 1 }{ 4{ x }_{ 0 } } }{ 2 } \ge \sqrt { \frac { 1 }{ 4 } }$

${ ({ x }_{ 0 }+\frac { 1 }{ 4{ x }_{ 0 } } ) }^{ 3 }\ge 1$

Finally $\boxed{{A}_{min}=\frac{4}{3}}$