Normal lines to a parabola, and areas bounded by them

Question

This is the question:

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What I have done:

(a) Show that the equation of the normal to the parabola at a point $(x_0,y_0)$ is $y = {-1\over 2kx_0} + kx_0^2 + {1\over 2k}$

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

$$ f( x_{0}) = kx_{0}^2 $$

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

$$ f'(x_{0}) = 2kx_0 $$

$$ Normal = -1/m $$

$$ m= {-1\over 2kx_0} $$

$$ y-y_1 = m (x-x_1) $$

$$ y-kx_0^2 = {-1\over 2kx_0}(x-x_0) $$

$$ y = {-1\over 2kx_0} + kx_0^2 + {1\over 2k} $$

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

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

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

$$ m = \sqrt{2} $$

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

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

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

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

$$ y-y_1 = m (x-x_1) $$

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

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

(c) Find the equation of the normal that produces the smallest area between itself and the parabola, and find this area.

This is the part where I'm stuck on.. how do I know which line will produce the smallest area?

Answer 1

Part (a) looks right to me.
For part (b), you need to find the other point $(x_1,y_1)$ as a function of $x_0$, and then find where $\frac{dx_1}{dx_0}=0$. You have started with the answer, and checked that it obeys part (a), but haven't explained why this one has the lowest point.
For part (c), once you have $x_1$ from part (b) and the line from part (a), you can calculate the area $A=\int_{x_1}^{x_0}(y-kx^2)dx$. Then $dA/dx_0=0$ for the minimal area.

Answer 2

By symmetry, we can assume $x_o > 0$. Thus we first find the intersections of the normal line at $(x_0,kx_0^2)$ and the parabola $y = kx^2$. We have: $kx^2 = \dfrac{-x}{2kx_0}+kx_0^2+\dfrac{1}{2k}$. This gives $2$ solutions: $x = x_0, -x_0-\dfrac{1}{2k^2x_0}$. Thus the area between the normal line at $(x_0,kx_0^2)$ and the parabola is: $\displaystyle \int_{-x_0 - \frac{1}{2k^2x_0}}^{x_0} \left(\dfrac{-x}{2kx_0}+kx_0^2 + \dfrac{1}{2k}-kx^2\right)dx$. From this you have a cubic polynomial in $x_0$, and you can use derivative to finish.