Question

99 Points Show your work:

Answer 1

Let's consider a normal line that is perpendicular to the curve in the first quadrant at the abcissa $a$. Being locally perpendicular to the curve, its slope is perpendicular to the curve's local slope, which is its derivative at $a$. The derivative of $x^2$ is $2x$ hence the curve's local slope at that point is $2a$. As a result, the normal line's slope is $-\frac{1}{2a}$. The normal line's curve's equation is thus of the form $-\frac{1}{2a}x+b$. What is more, its value at $a$ is $a^2$ hence $a^2=-\frac{1}{2a}a+b$ thus $b=x_0^2+\frac{1}2$. Hence the equation of the normal line is $\boxed{y=-\frac{1}{2a}x+a^2+\frac{1}2}$

Let's proceed to finding where it intercept the curve in the second quadrant. We have to solve $x^2=-\frac{1}{2a}x+a^2+\frac{1}2$. This equation's discriminant is $\Delta=\frac{1}{4a^2}+4\left(a^2+\frac{1}2\right)\ \textgreater \ 0$. This yields two solutions $x=\dfrac{-\frac{1}{2a}\pm\sqrt{\Delta}}{2}$, one of which is negative (and hence in the second quadrant), that is $\boxed{x=\dfrac{-\frac{1}{2a}-\sqrt{\frac{1}{4a^2}+4\left(a^2+\frac{1}2\right)}}{2}}$.

The "extreme normal line" is the one for which that absissa is the greatest (i.e. the closest to $0$ since they're negative). Let us this derive $x(a)$ with respect to $a$ : $x'(a)=-\frac{1}2\left[-\frac{1}{a^2}+\dfrac{-\frac{8a^3}+8a}{2\sqrt{\frac{1}{4a^2}+4(a^2+\frac{1}2)}}\right]$. $x'(a)=0\Leftrightarrow 2\sqrt{\underset{\frac{(4a^2+1)^2}{4a^2}}{\underbrace{\frac{1}{4a^2}+4(a^2+\frac{1}2)}}}+\frac{1}{8a^3}-8a=0 \Leftrightarrow 1-64a^4+8(4a^2+1)=0$.
Setting $A=a^2$, this leads to $A^2-\frac{A}2-\frac{9}{64}=0$, whose positive root is $A=\frac{1}4+\frac{\sqrt{13}}8$. This yields $\boxed{a=\sqrt{\frac{1}4+\frac{\sqrt{13}}8}}$.

The equation of the extreme normal line is therefore $\boxed{y=-\frac{x}{2a}+a^2+\frac{1}2,a=\sqrt{\frac{1}4+\frac{\sqrt{13}}8}}}$

The second question is extremely similar to this one, so I'll just show you the right track :

Let's call $d(a)=\dfrac{-\frac{1}{2a}-\sqrt{\frac{1}{4a^2}+4\left(a^2+\frac{1}2\right)}}{2}$ the abcissa of the intersection of a normal ine at $a$ and the curve, which we have computed before. The area between the normal line and the curve is $\int_{d(a)}^ax^2 dx=\left[\frac{x^3}3\right]_{d(a)}^a=\frac{1}3\left(a^3-d^3(a)\right)$. Now, derive this with respect to $a$ in order to find its extrema. You will some $a_0$ at which it is minimum, and hence the equation of the normal line that traps the least area between itself and the curve will be $y=-\frac{x}{2a_0}+a_0^2+\frac{1}2$

Don't hesitate to ask if you have any question :-) it's a very interesting problem !

Answer 2

Hope this helps! -John