Find the shortest distance between the point $(p,0)$, where $p> 0$, and the parabola $y^2=4ax$, where $a>0$, in the different cases that arise according to the value of $p/a$.
[You may wish to use the parametric coordinates $(at^2, 2at)$ of points on the parabola]Hence find the shortest distance between the circle $(x-p)^2 +y^2 = b^2$, where $p>0$ and $b>0$ and the parabola $y^2=4ax$, where $a>0$, in the different cases that arise according to the values of $p, a, b$.
I don't have much clue here – as usual I started off to find $\dfrac {dy}{dx}$ of the parametric equation = $\dfrac 1 t$
So the normal grad is $-t$ and it passes through $(p, 0)$.
Or we have two parallel lines consisting of
$$y=\dfrac 1 t x – \dfrac 1 t p$$
$$y=\dfrac 1 t x – at$$
and find the shortest distance between them.
I know
$$-t = \dfrac {-2at}{p -at^2} $$ and I tried to find t in terms of p and a,
But the distance between them using the equation
$$d = \dfrac{\left| d2-d1 \right|}{\sqrt {m^2 + 1}}$$
Gets really messy and does not seem to give what the solution paper says:
Shortest distance is $p$ if $p < 2$ and is $2\sqrt{a(p-a)}$ if $\dfrac p a \geq 2$.
Is there a quick way of doing this?
Further, in the previous part of this question
The line $L$ has equation $y=mx+c$, where $m>0$ and $c>0$. Show that, in the case $mc>a>0$, the shortest distance between L and the parabola $y^2 = 4ax$
$$\dfrac {mc-a}{m\sqrt{m^2 +1}}$$
I solved it using that $d$ equation I found from the Wikipedia. It is not given in the formula book so I think I am supposed to find an alternative way of solving it. (Or derive it on my own)
Equating two normal equations and comparing it to the original equation gives a point $(\dfrac a {m^2}, \dfrac {2a} {m})$
nNw $$d = \sqrt {(\dfrac a {m^2} – x)^2+{(\dfrac {2a} {m} – (mx +c)})^2}$$
Differentiating to find the maximum value gives :
$$=> x = \dfrac {c-\dfrac {2a}{m} – \dfrac {a}{m^2}}{m+1}$$
Again it becomes very messy when I put this $x$ back in to the equation to find $d$.
Is there a better way of doing this?
I know this block of text here may put off some people, but I wanted to show my working as much as possible.
Many thanks in advance.
Best Answer
Minimize for $t$
$$d^2(t)=(at^2-p)^2+(2at)^2.$$
$$(d^2(t))'=2(at^2-p)2at+2(2at)2a=0.$$
Then $$t=0\lor\left(p\ge2a\land at^2=p-2a\right),$$
The first case gives
$$d^2=p^2,$$ and the second $$d^2=4a(p-a).$$
Take the smallest of the two.