Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In the discriminant test of conic sections(rotations), why we're checking with $B^2-4AC$. How $B^2-4AC=B'^2-4A'C'$, where $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ is changed to $A'x^2+C'y^2+D'x+E'y+F'=0$ using rotations by angle alpha.

share|improve this question
    
You need to do the replacements of $x,y$ by the rotation, and then find the new $A',B'$ etc, and just compute $B'^2-A'C'$. If you have rotated so as to eliminate the $xy$ term, the $B'$ will be zero here. –  coffeemath Oct 22 '12 at 11:35
add comment

1 Answer 1

We can find from here, the condition for the general 2nd degree equation to represent a pair of straight lines.

If that condition is not satisfied we can apply the following logic.

A conic is the locus of a point $(x,y)$ which maintains constant ratio(called eccentricity$(e)$) of the distances from a fixed point(called focus $(h,k)$ (say),) and a fixed line(called directrix) $lx+my+n=0$(say).

So, $$e=\frac{(x-h)^2+(y-k)^2}{\frac{lh+mk+n}{\sqrt{l^2+m^2}}}$$

On squaring and rearrangement we get,

$$\{m^2+(1-e^2)l^2\}x^2-2lme^2xy+\{l^2+(1-e)^2m^2\}y^2+(..)x+(..)y+(..)=0$$

Comparing with the original equation, $A=m^2+(1-e^2)l^2,B=-2lme^2, C=l^2+(1-e)^2m^2$

So, $B^2-4AC=4(e^2-1)(l^2+m^2)^2$

Now,

for ellipse $0\le e<1, B^2-4AC=4(e^2-1)(l^2+m^2)^2<0$

for parabola $e=1, B^2-4AC=4(e^2-1)(l^2+m^2)^2=0$

for hyperbola $e>1, B^2-4AC=4(e^2-1)(l^2+m^2)^2>0$

for circle, $A=C\implies m^2+(1-e^2)l^2=l^2+(1-e)^2m^2\implies e=0$,the $xy$ must be absent in the general equation and the focus co0incides with the centre.


Using Rotation of axes,

$A'=A\cos^2\alpha+B\sin\alpha \cos\alpha +C\sin^2\alpha$

$\implies 2A'=(A+C)+(A-C)\cos2\alpha+B\sin2\alpha$

$B'=B(\cos^2\alpha-\sin^2\alpha)-2(A-C)\sin\alpha \cos\alpha=B\cos2\alpha-(A-C)\sin2\alpha$

$C'=A\sin^2\alpha-B\sin\alpha \cos\alpha +C\cos^2\alpha$

$\implies 2C'=A+C-\{B\sin2\alpha+(A-C)cos2\alpha\}$

$2A'\cdot 2C'-B'^2$ $=(A+C)^2-\{B\sin2\alpha+(A-C)cos2\alpha\}^2-\{B\cos2\alpha-(A-C)\sin2\alpha\}^2$ $=(A+C)^2-B^2\{\sin^22\alpha+\cos^22\alpha\}-2B(A-C)\{2\sin2\alpha\cos2\alpha-2\sin2\alpha\cos2\alpha\}-(A-C)^2\{\sin^22\alpha+\cos^22\alpha\}$ $=(A+C)^2-(A-C)^2-B^2$

$$\implies 4A'C'-B'^2=4AC-B^2$$ (This can be established by eliminating $\sin2\alpha, \cos2\alpha$ from the three equations.)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.