Finding Intersection Points Between Circle & Line

[Lancelot1]

Hello,

I wish to verify that the following pair ofcurves meet orthogonally.

\[x^{2}+y^{2}=4\]

and

\[x^{2}=3y^{2}\]

I recognize that the first is a circle, and the second contains 2 lines (y=1/3*x and y=-1/3*x).

I thought to get an implicit derivative of the circle, and to compare it to the line, showing that the slopes multiplication is -1. It did not work. How do I find the intersection points between the line and the circle? How to I proceed if the implicit derivative is -y/x ?

Thank you in advance.

 

[skeeter]

$x^2+y^2=4$

$x^2=3y^2$

sub ...

$3y^2+y^2=4 \implies y = \pm 1 \implies x = \pm \sqrt{3}$

four points of intersection ...

$(1,\sqrt{3}), \, (1,-\sqrt{3}), \, (-1,\sqrt{3}), \, (-1,-\sqrt{3})$

note ...

$x^2+y^2=4 \implies \dfrac{dy}{dx} = -\dfrac{x}{y}$

$x^2 = 3y^2 \implies \dfrac{dy}{dx} = \dfrac{x}{3y}$

$-\dfrac{x}{y} \cdot \dfrac{x}{3y} = -\dfrac{x^2}{3y^2} = -1$

product of the slopes is $-1$ ...

 

[MarkFL]

In general, suppose we have the following two families of curves:

$$x^2+y^2=r^2\implies \d{y}{x}=-\frac{x}{y}$$

$$x^2=ky^2\implies \d{y}{x}=\frac{x}{ky}$$

At the points of intersection of the two families, we find:

$$\left(x^2,y^2\right)=\left(\frac{kr^2}{k+1},\frac{r^2}{k+1}\right)$$

The product $P$ of the slopes of the two families is:

$$P=-\frac{x^2}{ky^2}$$

And so at the points of intersection, we have:

$$P=-\frac{\dfrac{kr^2}{k+1}}{k\dfrac{r^2}{k+1}}=-1$$

And so we conclude the families are orthogonal trajectories. :D

 

[Lancelot1]

Thank you both !

skeeter, I think you got the points in the wrong order (y,x), but I get the idea, thank you ! :D

 

[skeeter]

... skeeter, I think you got the points in the wrong order (y,x), but I get the idea, thank you ! :D

sorry ... discalcula error.