Solve the given parametric equation

[chwala]

Homework Statement

See attached

Relevant Equations

differentiation

For part (a) i have two approaches;
We can have,

$\dfrac{dy}{dx}=\dfrac{dy}{dt}\cdot\dfrac{dt}{dx}$
$\dfrac{dy}{dx}=-\dfrac{2}{x^2}$
$\dfrac{dy}{dx}\left[x=\frac{1}{p}\right]=-2p^2$

Therefore,
$p(y-2p)=-2p^3x+2p^2$
$py=-2p^3x+4p^2$
$y=-2p^2x+4p$The other approach to this is;
since $y=2t$ and $x=\dfrac{1}{t}$ then we shall have the cartesian equation, $xy=2.$ It follows that
$y+x\dfrac{dy}{dx}=0$
$\dfrac{dy}{dx}=-\dfrac{y}{x}=-2p^2$

The other steps shall follow as previously shown.

For part (b),
We have
$y=-2p^2x+4p$
at $x$ axis, $y=0$ therefore
$2p^2x=4p$
$px=2$
$x=\dfrac{2}{p}$
The co ordinates at $A\left[\dfrac{2}{p},0\right]$
$PA=\sqrt{\left[\dfrac{2}{p}-\dfrac{1}{p}\right]^2+[2p-0]^2}=\dfrac{\sqrt{4p^4+1}}{p}$

Let the gradient of the normal be given by $m$,
$-2p^2\cdot m=-1$
$m=\dfrac{1}{2p^2}$
therefore we shall have for $PB$
$\dfrac{y-2p}{px-1}=\dfrac{1}{2p^2}$
$2p^3(y-2p)=px-1$
$-4p^4=px-1$
$x=\dfrac{1-4p^4}{p}$

The co ordinates at $B\left[\dfrac{1-4p^4}{p},0\right]$

$PB=\sqrt{\left[\dfrac{1-4p^4}{p}-\dfrac{1}{p}\right]^2+[2p-0]^2}=\sqrt{16p^6+4p^2}=\sqrt{4p^2(4p^4+1)}=2p\cdot\sqrt{4p^4+1}$

Therefore, $PA:PB=\dfrac{1}{p}:2p=1:2p^2$

Of course this was easy...i always like seeking different ways if any to solve this hence my posts...Cheers guys

 

[pasmith]

I don't know if you can do something cunning using the fact that the curve is a conic section (hyperbola).

 

[Delta2]

I spotted one small typo, you have omitted a $p$ factor in the equation following the

Homework Statement:: See attached
Relevant Equations:: differentiation

therefore we shall have for PB

The correct should be $$\frac{p(y-2p)}{px-1}=\frac{1}{2p^2}$$. The rest of solution remains correct cause you bring back the p factor in the following lines.

 

[Maarten Havinga]

You can solve (b) by taking $x'=dx/dt$ and $y'=dy/dt$ at P. The length ratio of PA to PB will be $|y'/x'|$ due to $(P,A,X)$ being congruent to $(B,A,P)$. This evaluates to 2 : $p^{-2}$ and multiply by $p^2$ to get the right answer! :-)