Chapter 10.5, Problem 55AYU

To determine

To prove: $B²-4AC$ is invariant.

Answer to Problem 55AYU

$B²-4AC$ is invariant.

Explanation of Solution

Given:

The Equations,

$A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

And

$Ax{\text{'}}^2+Bx\text{'}y\text{'}+Cy{\text{'}}^2+Dx\text{'}+Ey\text{'}+F\text{'}=0$.

Proof:

The Equation $Ax²+Bxy+Cy²+Dx+Ey+F=0,B\ne 0$.

Let us consider,

$B^{\prime }²-4A^{\prime }{C}^{\prime }=(B\left({\cos }^2\theta -{\sin }^2\theta \right)+2\left(C-A\right)\sin \theta \cos \theta )²-4*\left(A{\cos }^2\theta +B\sin \theta \cos \theta +C{\sin }^2\theta \right)(A\sin ²\theta -B\sin \theta \cos \theta +C\cos ²\theta )$

$B^{\prime }²=B²\left({\cos }^2\theta -{\sin }^2\theta \right)²+4\left(A-C\right)\text{Sin}\theta \cos \theta \left(B^2{\left({\cos }^2\theta -{\sin }^2\theta \right)}^2\right)+4\left(A-C\right)²\sin ²\theta \cos ²\theta$

$B^{\prime }²=B²\cos ²2\theta -2B\left(A-C\right)\sin 2\theta \cos 2\theta +{\left(A-C\right)}^2{\sin }^{2}2\theta$

$4A^{\prime }{C}^{\prime }=4\left(A{\cos }^2\theta +B\sin \theta \cos \theta +C{\sin }^2\theta \right)\left(A{\sin }^2\theta -B\sin \theta \cos \theta +C{\cos }^2\theta \right)$

$4A^{\prime }{C}^{\prime }=(A\left(1+\cos 2\theta \right)+B\sin 2\theta +\left(1-\cos 2\theta \right)\left(A\left(1-\cos 2\theta \right)-B\sin 2\theta +C\left(1+\cos 2\theta \right)\right)$

$4A^{\prime }{C}^{\prime }=\left(A+C\right)+B\sin 2\theta +\left(A-C\right)\cos 2\theta )(\left(A+C\right)-\left(B\sin 2\theta +\left(A-C\right)\cos 2\theta \right)$

$4A^{\prime }{C}^{\prime }={\left(A+C\right)}^2-\left(B\sin 2\theta +\left(A-C\right)\cos 2\theta \right)²$

$4A^{\prime }{C}^{\prime }={\left(A+C\right)}^2-\left(B^2{\sin }^{2}2\theta +{\left(A-C\right)}^2{\cos }^{2}2\theta +2B\sin 2\theta \left(A-C\right)\cos 2\theta \right)$

$4A^{\prime }{C}^{\prime }={\left(A+C\right)}^2-\left(B^2{\sin }^{2}2\theta +{\left(A-C\right)}^2{\cos }^{2}2\theta +2B\left(A-C\right)\sin 2\theta \cos 2\theta \right)$

$B^{\prime }²-4A^{\prime }{C}^{\prime }=B^2\left({\sin }^{2}2\theta +{\cos }^{2}2\theta \right)+{\left(A-C\right)}^2\left({\cos }^{2}2\theta +{\sin }^{2}2\theta \right)-\left(A+C\right)²$

$B^{\prime }²-4A^{\prime }{C}^{\prime }=B^2\left({\sin }^{2}2\theta +{\cos }^{2}2\theta \right)+{\left(A-C\right)}^2\left({\cos }^{2}2\theta +{\sin }^{2}2\theta \right)-\left(A+C\right)²$

$B^{\prime }²-4A^{\prime }{C}^{\prime }=B²+A²+C²+2AC-A²-C²+2AC$

$B^{\prime }²-4A^{\prime }{C}^{\prime }=B²-4AC$

Hence $B²-4AC$ is invariant.