Prove that PA=2BP in the problem involving parametric equations

[chwala]

Homework Statement

Let P be a point on the curve ##x=t^2, y=\dfrac{1}{t}##. If the tangents to the curve at P meets the x- and y-axes at A and B respectively, prove that PA=2BP.

Relevant Equations

Parametric equations

My take;

##\dfrac{dy}{dx}=\dfrac{-1}{t^2}⋅\dfrac{1}{2t}=\dfrac{-1}{2t^3}##

The equation of the tangent line AB is given by;

##y-\dfrac{1}{t}=\dfrac{-1}{2t^3}(x-t^2)##

##ty=\dfrac{-1}{2t^2}(x-t^2)+1##

At point A, ##(x,y)=(3t^2,0)##

At point B, ##(x,y)=(0,1.5t)##

PA=##\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}##

=##\sqrt{(t^2-3t^2)^2+(\dfrac{1}{t}-0)^2}##
...

= ##\dfrac{\sqrt{4t^6+1}}{t}##


BP=##\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}##

=##\sqrt{(t^2-0)^2+(\dfrac{1}{t}-\dfrac{1.5}{t})^2}##
...

= ##\dfrac{\sqrt{4t^6+1}}{2t}##

it follows that ##\dfrac{PA}{BP}##=##\dfrac{\sqrt{4t^6+1}}{t}\div\dfrac{\sqrt{4t^6+1}}{2t}=2##

##⇒PA=2BP##thus proved.

insight welcome or other approach.

 

[Steve4Physics]

insight welcome or other approach.

You can use tools such as https://www.geogebra.org/m/cAsHbXEU to plot the curve.

Draw the tangent, AB, at some arbitrary point P.

Drop a perpendicular from P ##(t^2, \frac 1t)##.to meet the (say) x-axis at Q. Note that Q has coordinates ##(t^2, 0)##

You have already established that point A is ##(3t^2,0)##. You can now complete the proof using proportionality (the triangle proportionality theorem).

 

[pasmith]

Parametrize the tangent as [tex]
(t^2, t^{-1}) + \lambda(2t, -t^{-2}).[/tex] From here it is straightforward to find the intercepts at [itex]A = (3t^2,0)[/itex] and [itex]B = (0, \frac32t^{-1})[/itex] and then compute [tex]
\begin{split}
|PA| &= \sqrt{ (2t^2)^2 + (-t^{-1})^2 } = \sqrt{ 4t^4 + t^{-2}} \\
|PB| &= \sqrt { (-t^2)^2 + \left(\tfrac12 t^{-1}\right)^2 } = \sqrt{ t^4 + \tfrac14t^{-2}} \\
&= \tfrac12 |PA|
\end{split}[/tex]