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If $ \left| a\sin^2\theta+b\sin\theta\cos\theta+c\cos^2\theta-\dfrac{(a+c)}{2} \right|=\dfrac{k}{2}$, then find $k^2.$

My Attempt \begin{align} \pm k &= 2a\sin^2 \theta+2b\sin \theta \cos \theta+2c\cos^2 \theta-(a+c) \\ &= a[1-\cos 2\theta]+b\sin 2\theta+c[1+\cos 2\theta]-(a+c) \\ &= a\left[ 1-\frac{1-\tan^2 \theta}{1+\tan^2 \theta} \right]+ b\frac{2\tan \theta}{1+\tan^2 \theta}+ c\left[ 1+\frac{1-\tan^2 \theta}{1+\tan^2 \theta} \right]-(a+c)\\ \pm k[1+\tan^2 \theta] &= 2a\tan^2 \theta+2b\tan \theta+2c-(a+c)[1+\tan^2 \theta] \\ 0 &= \tan^2 \theta [a-c\mp k]+2b\tan \theta+(c-a\mp k) \\ \Delta &= 4b^2-4(a-c\mp k)(c-a\mp k) \\ &= 4b^2-4[\mp k+(a-c)][\mp k-(a-c)] \\ &= 4b^2-4[k^2-(a-c)^2] \\ & \ge 0 \\ b^2+(a-c)^2 & \ge k^2 \end{align}

How do I prove that $k^2=b^2+(a-c)^2$ ?

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    $\begingroup$ $$2a\sin^2t+2b\sin t\cos t+2c\cos^2t-(a+c)=\cos2t(c-a)+b\sin2t\le\sqrt{b^2+(c-a)^ 2}$$ I think you need minimum value of $k$ $\endgroup$ – lab bhattacharjee Nov 14 '18 at 7:22
  • $\begingroup$ You can not prove $k^2=b^2+(a-c)^2$, because for $\theta = 0$, $k^2=(c-a)^2$. $\endgroup$ – farruhota Nov 14 '18 at 10:52

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