Proving the Locus of a Moving Point is an Ellipse: Coordinate Geometry Problem

[maverick280857]

Hello...

Here is a problem in coordinate geometry, in particular about the ellipse.

A point moves such that the sum of the squares of its distances from two intersecting straight lines is constant. Prove that its locus is an ellipse and find the eccentricity in terms of the angle between the straight lines.

My solution:

Without loss of generality we may assume the two straight lines to be $y = 0$ and $y = mx$ where $m = \tan\phi$ ($\phi$ is the angle between the lines). Their point of intersection is thus the origin O(0,0).

Let the point whose locus is to be found be $P(\alpha,\beta)$. The constraint on P is then,

$$\beta^2 + \frac{(m\alpha - \beta)^2}{m^2+1} = k^2$$

where k is some constant ($k\epsilonR$)

This after some rearranging and replacing $(\alpha,\beta)$ with with general coordinates $(x,y)$ yields
$m^2x^2 - 2mxy + y^2(m^2+2) - k^2(1+m^2) = 0$
which when compared with the general second degree equation,
$Ax^2 + 2Hxy + By^2 + 2gx + 2fy + c = 0$
does turn out to be an ellipse.

However it is not in the standard form, so finding its eccentricity is not as easy. Now I understand that by rotating the coordinate axes we can bring the equation into such a form by a suitable choice of the rotation angle which causes the cross term (H) to disappear. However, I want to know if there is some other way out to find the eccentricity (or more generally to do this problem).

I would be grateful if someone could offer some ideas.

Thanks and cheers
Vivek

 

[maverick280857]

After posting this, I realized this probably isn't the right place for this post. For the moderator(s): if you think, could you please shift it to the right place. Sorry for the inconvenience...

 

[maverick280857]

Someone please help!

 

[arildno]

What's wrong with writing it in standard form??

 

[dextercioby]

I don't see any other way to extract the excentricity,other than knowing the semiaxes...You can do that simply applying the theory of conics and the set of linear transformations which bring the conic to a known form,in this case an ellipse.

Daniel.

 

[maverick280857]

Okay thanks for your replies. Suppose that I live in a very weird world and I'm supposed to show that I can do this "smartly" but not in a lengthy way. How would I do it?

I know I can rotate the axes and do what I've said I can in the first post to get it in the standard form. But I was wondering if there's some other way out. Anyway thanks...

Cheers
vivek

 

[dextercioby]

How about if i told you that the equation:
$$m^{2}x^{2}-2mxy+y^{2}(m^{2}+2)-k^{2}(m^{2}+1)$$(1)

is the EQUATION OF A CIRCLE...

The equation (1) can be written:
$$(mx-y)^{2}+[y\sqrt{m^{2}+1}]^{2} =[k\sqrt{m^{2}+1}]^{2}$$ (2)

and making the rotation & the notation:
$$x'=:mx-y$$ (3)
$$y'=y\sqrt{m^{2}+1}$$ (4)
$$R=:k\sqrt{m^{2}+1}$$ (5)

is exactly the equation of a circle:
$$x'^{2}+y'^{2}=R^{2}$$ (6)

If this outcome is not correct,then it's your fault for providing an incorrect quadratic form...

Daniel.

 

[maverick280857]

I do not know what you are trying to imply. The question and my working are before you. Do you believe the question is wrong?

 

[dextercioby]

I don't know that.I've just shown you that the equation u came up with when substituting alpha & beta wiht x & y is the equation of a circle,and not of an ellipse.

Daniel.

 

[chronon]

Without loss of generality we may assume the two straight lines to be $y = 0$ and $y = mx$ where $m = \tan\phi$ ($\phi$ is the angle between the lines).

You could try assuming $y = mx$ and $y = -mx$ where $m = \tan\phi/2$

$$x'=:mx-y$$ (3)
$$y'=y\sqrt{m^{2}+1}$$ (4)

This is squashing the coordinates, and so will transform a circle into an ellipse

 

[dextercioby]

You could try assuming $y = mx$ and $y = -mx$ where $m = \tan\phi/2$

This is squashing the coordinates, and so will transform a circle into an ellipse

Ups,you made me realize that what i had was not a genuine rotation and indeed i squashed "y" and so the initial ellipse became a circle...
Sorry.

I would have to stick to the initial advice,namely applying the theory of conics...

Daniel.

 

[maverick280857]

You could try assuming $y = mx$ and $y = -mx$ where $m = \tan\phi/2$

This is squashing the coordinates, and so will transform a circle into an ellipse

Yes but the problem is to extract the eccentricity from the equation without reducing it to a standard form via rotation. However, as the problem stands now, I do not think there is any other method than to do it the brute force way. Thanks for your help though.

Cheers
vivek