equidistant set

[renegade05]

Here is the question:

Show that the set of points equidistant from a circle and a line not passing through the circle is a parabola.

To make everything easier, i know i should let the parabola be symmetric about the y-axis and have the vertex at the origin. I know i should prob define the radius of the circle as r, and make it centered at (r+a, 0) and have a line y=-a.

After defining everything and setting it up like this i am stuck. Need some help.

Thanks!

 

[tkhunny]

I'm not entirely convinced this is the most general case.

Anyway, the distance from your line to your circle...

Points in the 2d universe look like this: (x,y)

Distance to the line is simply: y - (-a) = y+a

The distance to the circle is \(\displaystyle \sqrt{(x-0)^{2} + (y-a)^2} - r\)

The rest is algebra.

 

[Deleted member 4993]

Here is the question:

Show that the set of points equidistant from a circle and a line not passing through the circle is a parabola.

To make everything easier, i know i should let the parabola be symmetric about the y-axis and have the vertex at the origin. I know i should prob define the radius of the circle as r, and make it centered at (r+a, 0) and have a line y=-a.

After defining everything and setting it up like this i am stuck. Need some help.

Thanks!

Distance of a point from a circle does not make sense. Could that be "shortest distance"?

Most general definition of parabola is:

a plane curve generated by a point moving so that its distance from a fixed point is equal to its distance from a fixed line

or

the intersection of a right circular cone with a plane parallel to an element of the cone

 

[renegade05]

It doesn't say anything about shortest distance.

I don't think it matters? I am just looking for the equation of the parabola.

My teacher posted the solution.

The answer is: \(\displaystyle y=\frac{1}{4(r+a)}x^2\) which is the equation of a parabola.

I think i understand, the algebra is very annoying though, I think i keep making a sign error somewhere.

 

[Deleted member 4993]

It doesn't say anything about shortest distance.

I don't think it matters? I am just looking for the equation of the parabola.

My teacher posted the solution.

The answer is: \(\displaystyle y=\frac{1}{4(r+a)}x^2\) which is the equation of a parabola.

I think i understand, the algebra is very annoying though, I think i keep making a sign error somewhere.

So the distance from circle is the shortest distance or the "normal" distance - according to the answer.

Can you show that:

The locii of points whose distance from a fixed point (focus) is equal to its distance from a fixed line(directrix) is a parabola.

It will follow the same "construction".

 

[galactus]

Take a look at your given solution and take a look at the equation, \(\displaystyle x^{2}=4py\), for a concave up parabola. Where (0,p) is the coordinates of the focus and y=-p is the equation of the directrix line.

What if you solve this for y and let p=r+a?.

Though, you may want to derive \(\displaystyle x^{2}=4py\), which is not too involved.