Could anyone out there tell me how, via converting (#) to cartesian co-ordinates, I can show that (#) describes any conic section, depending on the values of the constants k and m?

$1/r=k\cos (\theta )+m$ ......(#)
(P.S. There are supposed to be only 3 conic sections: parabola, ellipse and hyperbola).
Thanks for any solutions.

I have an idea...

Put $x=r\cos \theta$ and $y=r\sin \theta$

$1/r=kx/r+m$

$1=kx+mr$

$r=\sqrt{x^2+y^2}$

$1=kx+m\sqrt{x^2+y^2}$

$m\sqrt{x^2+y^2}=l-kx$
Squaring and rearranging you obtain:

$m^2{x}^2+m^2{y}^2+2xk-x^{2}k-1=0$
The general equation of a circle is,
$x^2+y^2+2gx+2fy+c=0$

with centre $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$
love arun

When the coefficient of x² is different from that of y² the equation seems to resemble that of an ellipse. The general equation for an ellipse is:

(x-x_c )² /a + (y-y_c )² /b = 1 (a,b > 0)

where (x_c ,y_c ) is the centre of the ellipse. You can complete the square for x then divide throughout by the remaining constant to get your example in this form.

I'm not sure what a and b (constants) represent, having not done much work in this area. Maybe someone else could explain.

Jim

I think this site should answer it:

http://thesaurus.maths.org/dictionary/map/word/3474 .

Just plug the values of the coefficients in and the answer should drop out.

Regards,
Olof

Hi,

In answering the original question,

1/r = kx/r + m
1 - kx = mr
Squaring both sides we obtain
x² (m² -k² ) + y² (m² ) + 2kx = 1

Now let k = -m. This will give you the equation for a parabola with x intercept -1/(2m).

Let k = -mp (where p > 1). This will give you hyperbola with x intercepts +/- 1/((p-1)m)

Let k = -m/q (where q > 1). This will give you an ellipse.

Obviously, when making the above substitutions you will have to complete squares and do some algebraic manipulation.

In the above substitutions we can also let k = m, or k = mp or k = m/q and still get the desired conic sections, only that this time they will be reflected in the y axis.

regards
dimitri