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