This is the question:

[![enter image description here](https://i.stack.imgur.com/89JCI.png)](https://i.stack.imgur.com/89JCI.png)

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What I have done:

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> (a) Show that the equation of the normal to the parabola at a point $(x\_0,y\_0)$ is $y = {-1\over 2kx\_0} + kx\_0^2 + {1\over 2k}$
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$$ f(x) = kx^2 $$

$$ f( x\_{0}) = kx\_{0}^2 $$

$$ f'(x) = 2kx $$

$$ f'(x\_{0}) = 2kx\_0 $$

$$ Normal = -1/m $$

$$ m= {-1\over 2kx\_0} $$

$$ y-y\_1 = m (x-x\_1) $$

$$ y-kx\_0^2 = {-1\over 2kx\_0}(x-x\_0) $$

$$ y = {-1\over 2kx\_0} + kx\_0^2 + {1\over 2k} $$

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> > (b) Show that the equation of the normal line with the minimum y-coordinate is $ y = \frac{-\sqrt{2}}{2}x + {1\over k} $
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$$ y = \frac{-\sqrt{2}}{2}x + {1\over k} $$

$$ {-1\over m} = \frac{-\sqrt{2}}{2} $$

$$ m = \sqrt{2} $$

$$ f'(x) = 2kx $$

$$ 2kx = \sqrt{2} $$

$$ x = {\sqrt{2}\over 2k} $$

$$ f({\sqrt{2}\over 2k}) = {1\over 2k}$$

$$ y-y\_1 = m (x-x\_1) $$

$$ y - {1\over 2k} = \frac{-\sqrt{2}}{2} (x-{\sqrt{2}\over 2k}) $$

$$ y = \frac{-\sqrt{2}}{2}x + {1\over k} $$

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> (c) Find the equation of the normal that produces the smallest area between itself and the parabola, and find this area. 
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This is the part where I'm stuck on.. how do I know which line will produce the smallest area?