Q.Show that the shortest distance of the point $(0, 8a)$ from the curve $ax^2 = y^3$ is $2a\sqrt{11}$.

Minimizing the squared distance from $(0,2)$ to a general point $(x,x^2+1)$ on the curve gives $x=\pm1/\sqrt2$.

A general point on the curve is $(x, x^2+1)$. Squared distance from $(0,2)$:

$D=x^2+(x^2+1-2)^2 = x^2+(x^2-1)^2$

Let $f(x)=x^2+(x^2-1)^2 = x^4-x^2+1$.

$f'(x)=4x^3-2x = 2x(2x^2-1)$

Setting $f'(x)=0$: $x=0$ or $x^2=\dfrac12\Rightarrow x=\pm\dfrac{1}{\sqrt2}$.

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Minimise the squared distance from $(3,7)$ to a general point $(x,x^2+7)$ on the curve using calculus. (OR: implicit differentiation of $ay^2=x^3$ to get the slope at the given point.)

Main part. A general point on $y=x^2+7$ is $(x,\,x^2+7)$. Squared distance from $(3,7)$:

$D^2 = (x-3)^2 + (x^2+7-7)^2 = (x-3)^2+x^4$

Let $f(x) = x^4+(x-3)^2 = x^4+x^2-6x+9$.

$f'(x) = 4x^3+2x-6 = 2(2x^3+x-3)$

Setting $f'(x)=0$: $2x^3+x-3=0$. Testing $x=1$: $2+1-3=0$ ✓, so $x=1$ is a root.

Factor: $2x^3+x-3 = (x-1)(2x^2+2x+3)$. The quadratic $2x^2+2x+3$ has discriminant $4-24=-20<0$ (no real roots), so $x=1$ is the only critical point.

$f''(x) = 12x^2+2 > 0 \text{ for all } x \;\Rightarrow\; x=1 \text{ is a minimum}$

$f(1) = 1+1-6+9 = 5 \;\Rightarrow\; D=\sqrt5$

(At $x=1$, the point is $(1,8)$; distance from $(3,7)$ is $\sqrt{(1-3)^2+(8-7)^2}=\sqrt{4+1}=\sqrt5$.)

OR. $ay^2=x^3$. Differentiating implicitly:

$2ay\frac{dy}{dx} = 3x^2 \;\Rightarrow\; \frac{dy}{dx} = \frac{3x^2}{2ay}$

At $(am^2,am^3)$: …

Express the squared distance from a general point on the curve to $(0,8a)$ as a function of $y$ alone, minimize it via calculus, and evaluate at the minimizing $y$.

Curve: $ax^2=y^3$, i.e. $x^2=\dfrac{y^3}{a}$. Point: $P=(0,8a)$.

For a point $(x,y)$ on the curve, squared distance to $P$:

$D^2 = x^2+(y-8a)^2 = \dfrac{y^3}{a}+(y-8a)^2$

Minimize with respect to $y$:

$\dfrac{d(D^2)}{dy} = \dfrac{3y^2}{a}+2(y-8a) = 0$

Multiply by $a$: $3y^2+2ay-16a^2=0$

$y = \dfrac{-2a\pm\sqrt{4a^2+192a^2}}{6} = \dfrac{-2a\pm\sqrt{196a^2}}{6} = \dfrac{-2a\pm14a}{6}$

So $y=2a$ or $y=-\dfrac{8a}{3}$.

Since the curve requires $y^3=ax^2\ge0$ (for $a>0$), $y$ must be $\ge0$; so $y=-\dfrac{8a}{3}$ is rejected, leaving $y=2a$.

At $y=2a$: $x^2 = \dfrac{(2a)^3}{a} = \dfrac{8a^3}{a} = 8a^2$

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Main: the helicopter is nearest at $(1,8)$, distance $\sqrt5$. OR: tangent $x-ty+at^2=0$, normal $tx+y-2at-at^3=0$.

Main part. A point on $y=x^2+7$ is $(x,x^2+7)$; distance from soldier $(3,7)$:

$D^2=(x-3)^2+(x^2+7-7)^2=(x-3)^2+x^4.$

Let $f(x)=x^4+(x-3)^2$. Then

$f'(x)=4x^3+2(x-3)=4x^3+2x-6=2(2x^3+x-3)=2(x-1)(2x^2+2x+3).$

Since $2x^2+2x+3$ has discriminant $4-24<0$ (no real root), the only critical point is $x=1$. As $f''(x)=12x^2+2>0$, this is a minimum.

Nearest point $(1,\,1^2+7)=(1,8)$ and

$D=\sqrt{(1-3)^2+1^4}=\sqrt{4+1}=\sqrt5.$

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Write a general point on the parabola in terms of y, minimize the squared distance to (2,-8) using calculus.

Any point on $y^2=4x$ can be written as $\left(\dfrac{y^2}4,\,y\right)$.

Squared distance to $(2,-8)$: $D(y) = \left(\dfrac{y^2}4-2\right)^2+(y+8)^2$

$\dfrac{dD}{dy} = 2\left(\dfrac{y^2}4-2\right)\cdot\dfrac y2 + 2(y+8) = \dfrac{y^3}4-2y+2y+16 = \dfrac{y^3}4+16$ …