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Question Number 25002 by ajfour last updated on 30/Nov/17

Commented by ajfour last updated on 30/Nov/17

$$SolutiontoQ.24966$$

Answered by ajfour last updated on 30/Nov/17

$$x^2-11y^2-16xy-12x+6y+21=0$$$$Focus=?$$$${(x-a)}^2-11{(y-b)}^2-16(x-a)(y-b)+c=0$$$$\Rightarrow x^2-11y^2-16xy+(-2a+16b)x$$$$+(22b+16a)+(a^2-11b^2-16ab+c)=0$$$$\Rightarrow -2a+16b=-12...\left(i\right)$$$$22b+16a=6...\left(ii\right)$$$$a^2-11b^2-16ab+c=21...\left(iii\right)$$$$\Rightarrow a=\frac{6}{5},b=-\frac{3}{5},c=12$$$$so{(x-\frac{6}{5})}^2-11{(y+\frac{3}{5})}^2$$$$-16(x-\frac{6}{5})(y+\frac{3}{5})+12=0$$$$.....\left(a\right)$$$$letx-\frac{6}{5}=X\cos \theta -Y\sin \theta$$$$y+\frac{3}{5}=X\sin \theta +Y\cos \theta$$$$uponsubstitutionin\left(a\right):$$$$(\cos {}^2\theta -11\sin {}^2\theta -16\sin \theta \cos \theta )X^2$$$$-(-\sin {}^2\theta +11\cos {}^2\theta -16\sin \theta \cos \theta )Y^2$$$$+(-24\sin \theta \cos \theta +16\sin {}^2\theta -16\cos {}^2\theta )XY+12=0$$$$Ifwedesirecoeff.ofXYtobe$$$$zero\Rightarrow$$$$-24\tan \theta +16\tan {}^2\theta -16=0$$$$\Rightarrow \tan \theta =2,-\frac{1}{2}$$$$letuschoose\tan \theta =2$$$$then\sin \theta =\frac{2}{\sqrt{5}},\cos \theta =\frac{1}{\sqrt{5}}$$$$soequationofhyperbola$$$$becomes:$$$$-15X^2+5Y^2+12=0$$$$or\frac{X^2}{{\left(\frac{2\sqrt{5}}{5}\right)}^2}-\frac{Y^2}{{\left(\frac{2\sqrt{15}}{5}\right)}^2}=1$$$$e^2=1+\frac{B^2}{A^2}=1+{\left(\frac{\sqrt{15}}{\sqrt{5}}\right)}^2=4$$$$e=2$$$$focusat(x_1,y_1)$$$$x_1=\frac{6}{5}+Ae\cos \theta$$$$=\frac{6}{5}+\left(\frac{2\sqrt{5}}{5}\right)\times 2\times \frac{1}{\sqrt{5}}=2$$$$y_1=-\frac{3}{5}+Ae\sin \theta$$$$=-\frac{3}{5}+\left(\frac{2\sqrt{5}}{5}\right)\times 2\times \frac{2}{\sqrt{5}}=1$$$$hencefocusat(2,1).$$

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