Question

99 Points Show your work:

Answer 1

Hope this helps! -John

Answer 2

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
`-(1)/(2a)`. The normal line's curve's equation is thus of the form
`-(1)/(2a)x+b`. What is more, its value at
`a` is
`a^2` hence
`a^2=-(1)/(2a)a+b` thus
`b=x_0^2+\frac{1}2`. Hence the equation of the normal line is
`\boxed{y=-(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=-(1)/(2a)x+a^2+\frac{1}2`. This equation's discriminant is
`\Delta=(1)/(4a^2)+4\left(a^2+\frac{1}2\right)\ \textgreater \ 0`. This yields two solutions
`x=(-(1)/(2a)\pm√(\Delta))/(2)`, one of which is negative (and hence in the second quadrant), that is
`\boxed{x=\frac{-(1)/(2a)-\sqrt{(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[-(1)/(a^2)+\frac{-\frac{8a^3}+8a}{2\sqrt{(1)/(4a^2)+4(a^2+\frac{1}2)}}\right]`.
`x'(a)=0\Leftrightarrow 2\sqrt{\underset{((4a^2+1)^2)/(4a^2)}{\underbrace{(1)/(4a^2)+4(a^2+\frac{1}2)}}}+(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-(9)/(64)=0`, whose positive root is
`A=\frac{1}4+\frac{√(13)}8`. This yields
`\boxed{a=\sqrt{\frac{1}4+\frac{√(13)}8}}`.

The equation of the extreme normal line is therefore
`\boxed{y=-(x)/(2a)+a^2+\frac{1}2,a=\sqrt{\frac{1}4+\frac{√(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)=\frac{-(1)/(2a)-\sqrt{(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=-(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 !