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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 [itex]y = 0[/itex] and [itex]y = mx[/itex] where [itex]m = \tan\phi[/itex] ([itex]\phi[/itex] 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 [itex]P(\alpha,\beta)[/itex]. The constraint on P is then,

[tex]\beta^2 + \frac{(m\alpha - \beta)^2}{m^2+1} = k^2 [/tex]

where k is some constant ([itex]k\epsilonR[/itex])

This after some rearranging and replacing [itex](\alpha,\beta)[/itex] with with general coordinates [itex](x,y)[/itex] yields

[itex]m^2x^2 - 2mxy + y^2(m^2+2) - k^2(1+m^2) = 0[/itex]

which when compared with the general second degree equation,

[itex]Ax^2 + 2Hxy + By^2 + 2gx + 2fy + c = 0 [/itex]

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