Paranormal

The gradient of the curve at the point $P\; (kt,kt^2)$ is $2t$ , so the equation of the normal to the curve at that point is

$x + 2ty \; = \; kt + 2kt^3$

and this curve meets the parabola at the point $(ks,ks^2)$ where $\begin{aligned} ks + 2kts^2 & = \; kt + 2kt^3 \\ k(s-t)\big(1 + 2t(s+t)\big) & = \; 0 \end{aligned}$ so the point $Q\;(ks,ks^2)$ is given by $s = -t - \tfrac{1}{2t}$ . The $y$ -coordinate of $Q$ is thus $k\big(t + \tfrac{1}{2t}\big)^2 \; = \; k\big(t - \tfrac{1}{2t}\big)^2 + 2k$ and so the smallest possible $y$ -coordinate of $Q$ is $2k$ . Thus we deduce that $k = \boxed{10}$ .

This isn't really an elegant solution. Also I'm a bit of a LaTex noob so please forgive the formatting.

$y=\frac{x^2}{k}$

We shall first find the equation of the normal line at $x=p$ .

The gradient of the normal line is $-\frac{1}{dy/dx}|=-\frac{1}{2x/k}|_{x=p}=\frac{-k}{2p}$ .

We let $y_{tan}$ be the normal line.

$y_{tan}(x)=\frac{-k}{2p}x+c$

$y_{tan}(p)=\frac{p^2}{k}$

$\Rightarrow \frac{-k}{2p}p+c=\frac{p^2}{k}$

$\Rightarrow c=\frac{p^2}{k}+\frac{k}{2}$

$\Rightarrow y_{tan}=\frac{k^2-2p^2}{2k}-\frac{kx}{2p}$

$y_{tan}(x)$ and $y=\frac{x^2}{k}$ intersect at points $P(p,\frac{p^2}{k})$ and $Q(q,\frac{q^2}{k})$ so we solve $\frac{k^2-2p^2}{2k}-\frac{kq}{2p}=\frac{q^2}{k}$ for $q$ .

Thus $q=\frac{-(k^2+2p^2)}{2p}$ (or $q=p$ which is not desired in this case)

We want the minimum of $\frac{q^2}{k}$ to be 20. So we define the function $f=\frac{q^2}{k}=\frac{(k^2+2p^2)^2}{4kp^2}$

Next, we solve $\frac{df}{dp}=0$ for $p$ in terms of $k$ and substitute it back in to $f$ .

$\frac{df}{dp}=\frac{(2p^2+k^2)(2p^2-k^2)}{2p^3 k}$

$\frac{(2p^2+k^2)(2p^2-k^2)}{2p^3 k}=0$

$\Rightarrow 2p^2+k^2=0$ or $2p^2-k^2=0$

$\Rightarrow p^2=\frac{-k^2}{2}$ or $p^2=\frac{k^2}{2}$

$f|_{p^2=\frac{-k^2}{2}}=\frac{(k^2+2(\frac{-k^2}{2}))^2}{4k(\frac{-k^2}{2})}$

$=\frac{0^2}{4k(\frac{-k^2}{2})}$

$=0$

(this answer doesn't mean much since $p$ is supposed to be a real number anyways)

$f|_{p^2=\frac{k^2}{2}}$ $=\frac{(k^2+2(\frac{k^2}{2}))^2}{4k(\frac{k^2}{2})}$ $=2k$ .

If we equate $2k=20$ , we obtain the desired $\boxed{k=10}$ . And for completeness, $\frac{q^2}{10}=20$ when $p=\pm 5\sqrt{2}$ . $\frac{q^2}{10}>20$ for any other values of $p$ .

Let $P$ be any point on the parabola $y=\dfrac{x^2}{k}\implies P=\left(a,\dfrac{a^2}{k}\right)$ for some real $a$ .

The slope of the normal at point $P$ is given by $-\dfrac{1}{\left. \dfrac{\mathrm dy}{\mathrm dx}\right]_{x=a}}=-\dfrac{1}{\left.\dfrac{2x}{k}\right]_{x=a}}=-\dfrac{k}{2a}$ Using the Point-Slope form, the equation of line $PQ$ is, $y-\dfrac{a^2}{k}=-\dfrac{k}{2a}(x-a)$ $\implies \frac{kx}{2a}+y-\frac{a^2}{k}-\frac{k}{2}=0$ In order to find the intersection points ( $P$ and $Q$ ) of line $PQ$ and parabola $y=\dfrac{x^2}{k}$ , we need to solve the pair of equations, $\begin{cases} \dfrac{kx}{2a}+y-\dfrac{a^2}{k}-\dfrac{k}{2}=0 \\ y= \dfrac{x^2}{k} \end{cases}$ This leads us to the solutions $\begin{aligned} P&=\left(a,\dfrac{a^2}{k}\right) \\ \\ Q&=\left(-\frac{k^2}{2a}-a, \; \frac{k^3}{4a^2}+\frac{a^2}{k}+k\right) \end{aligned}$ So, we are given that $\text{min}\left(\dfrac{k^3}{4a^2}+\dfrac{a^2}{k}+k\right)=20$ . Differentiating with respect to $a$ and equating to $0$ , $\dfrac{\mathrm d}{\mathrm da}\left(\dfrac{k^3}{4a^2}+\dfrac{a^2}{k}+k\right)=0$ $\begin{aligned} &\implies -\frac{k^3}{2a^3}+\frac{2a}{k}&=0 \\ \\ &\implies k =\sqrt{2}a \end{aligned}$ Therefore , $a=\dfrac{k}{\sqrt{2}}$ is a critical point, and is in fact, the point of minima. So, when $a=\dfrac{k\sqrt{2}}{2}$ , the expression $\dfrac{k^3}{4a^2}+\dfrac{a^2}{k}+k$ is equal to $20$ . $\begin{aligned} &\implies \dfrac{k^3}{4\left (\dfrac{k\sqrt{2}}{2}\right )^2}+\dfrac{\left (\dfrac{k\sqrt{2}}{2}\right )^2}{k}+k=20 \\ \\ &\implies \frac{k}{2}+\frac{k}{2}+k=20 \\ \\ &\implies \boxed{k=10}\end{aligned}$

Let the coordinates of $P$ be $P(p, \cfrac{p^2}{k})$ and let the coordinates of $Q$ be $Q(q, \cfrac{q^2}{k})$ .

The slope of the tangent at $P$ on $y = \cfrac{1}{k}x^2$ is $\cfrac{2p}{k}$ and the slope of the normal at $P$ is $-\cfrac{k}{2p}$ , making the equation of the normal $y = -\cfrac{k}{2p}(x - p) + \cfrac{p^2}{k}$ .

Since $Q(q, \cfrac{q^2}{k})$ is on the normal as well, $\cfrac{q^2}{k} = -\cfrac{k}{2p}(q - p) + \cfrac{p^2}{k}$ , which solves to $q = -p - \cfrac{k^2}{2p}$ , so that the $y$ -coordinate of $Q$ is $\cfrac{q^2}{k} = \cfrac{p^2}{k} + k + \cfrac{k^3}{4p^2}$ .

By the AM-GM inequality, $\cfrac{p^2}{k} + \cfrac{k^3}{4p^2} \geq 2\sqrt{\cfrac{p^2}{k} \cdot \cfrac{k^3}{4p^2}} = k$ , so $\cfrac{p^2}{k} + k + \cfrac{k^3}{4p^2} \geq k + k = 2k$ , which means the lowest possible $y$ -coordinate of $Q$ is $2k$ .

Therefore, $2k = 20$ , and $k = \boxed{10}$ .

$y_{0}^2 = \dfrac{x_{0}^2}{k} \implies y'_{0} = \dfrac{2x_{0}}{k} \implies m_{PQ} = -\dfrac{k}{2x_{0}}$

At $P: \:\ y = -\dfrac{k}{2x_{0}}(x - x_{0}) + \dfrac{x_{0}^2}{k}$

At $Q: \:\ \:\ x_{1}^2 = -\dfrac{k}{2x_{0}}(x_{1} - x_{0}) + \dfrac{x_{0}^2}{k}$

$\implies (2x_{0})x_{1}^2 + k^2x_{1} - (k^2x_{0} + 2x_{0}^3) = 0 \implies x_{1} = \dfrac{k^2 \pm (k^2 + 4x_{0}^2}{4x_{0}}$

For $(+) \:\ x_{1} = x_{0} \therefore$ drop $(+)$ and choose $x_{1} = \dfrac{k^2}{4x_{0}^2} - x_{0} \implies$ $\dfrac{x_{1}^2}{k} = \dfrac{k^3}{4x_{0}^2} + k + \dfrac{x_{0}^2}{k}$

Let $g(x_{0}) = \dfrac{k^3}{4x_{0}^2} + k + \dfrac{x_{0}^2}{k} \implies$

$P:(x_{0},\dfrac{x_{0}^2}{k})$ and $Q:(-\dfrac{k^2}{2x_{0}} - x_{0},g(x_{0}))$ .

$g(x_{0}) = \dfrac{k^3}{4x_{0}^2} + k + \dfrac{x_{0}^2}{k} \implies g'(x_{0}) = \dfrac{4x_{0}^4 - k^4}{2x_{0}^3k} = 0$

$\implies x_{0} = \dfrac{k}{\sqrt{2}}$

and

$(-\dfrac{k}{\sqrt{2}} < x_{0} < \dfrac{k}{\sqrt{2}} \implies g' < 0)$ and $(x_{0} > \dfrac{k}{\sqrt{2}} \implies g' > 0) \implies$ min occurs at $x_{0} = \dfrac{k}{\sqrt{2}}$

$\implies g(\dfrac{k}{\sqrt{2}}) = 2k = 20 \implies k = \boxed{10}$