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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.

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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.

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If $d$ is the distance between the point $(p,0)$, where $p> 0$, and the parabola $y^2=4ax$,

$$d^2=(at^2-p)^2+(2at-0)^2=a^2t^4+p^2+2at^2(2a-p)$$

$$=\left(at^2+2a-p\right)^2+p^2-(2a-p)^2\ge p^2-(2a-p)^2=4a(p-a)$$

The equality occurs if $at^2=p-2a$

Clearly, $p-2a\ge\iff p\ge2a$


If $D$ is the distance between the circle $(x−p)^2+y^2=b^2,$ and the parabola $y^2=4ax$

$$D^2=(at^2-b\cos u-p)^2+(2at-b\sin t)^2=\cdots$$

But we need the application of calculus here

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  • $\begingroup$ Thank you.. nice to spot that factorisation in the second line. I don't think I can solve it like this during the exam... $\endgroup$ – zcahfg2 Jun 7 '15 at 14:08
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There are several ways to approach this problem.

The simplest is to note that the distance between two points, one of which is (p, 0), is given by $\sqrt{(x- p)^2+ y^2}$ which will be minimum when its square, $(x- p)^2+ y^2$ is minimum. Requiring that (x, y) be on the parabola $y^2= 4ax$ means that $x= \frac{y^2}{4a}$. So the quantity to be minimized is $(y^2/4a- p)^2+ y^2= y^4/(16a^2)- py^2/2a+ p^2+ y^2= y^4/(16a^2)- (p- 2a)y^2/2a+ p^2$

That is a fourth degree polynomial in y but there are no odd powers so letting $u= y^2$, we have $u^2/(16a^2)- (p- 2a)u+ p^2$. You can find the minimum value of that, and the u value that makes it minimum, by [b]completing the square[/b].

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