91x^2 +84y^2 −24xy+406x−392y+799=0 find the eccentricity,focus,length of major & minor axis,directrix & length of eccentric perpendicular


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Question Number 168367 by MdNafiz last updated on 09/Apr/22

$91 x^{2} + 84 y^{2} - 24 x y + 406 x - 392 y + 799 = 0$ $f i n d t h e e c c e n t r i c i t y, f o c u s, l e n g t h o f m a j o r & m i n o r a x i s, d i r e c t r i x & l e n g t h o f e c c e n t r i c p e r p e n d i c u l a r$
	Answered by mr W last updated on 10/Apr/22

	$91 x^{2} - 2 (12 y - 203) x + 84 y^{2} - 392 y + 799 = 0$ $Δ = {(12 y - 203)}^{2} - 91 (84 y^{2} - 392 y + 799) ⩾ 0$ $75 y^{2} - 308 y + 315 ⩽ 0$ $\frac{154 - \sqrt{91}}{75} ⩽ y ⩽ \frac{154 + \sqrt{91}}{75}$ $\Rightarrow y_{C} = \frac{154}{75}$ $84 y^{2} - 2 (12 x + 196) y + 91 x^{2} + 406 x + 799 = 0$ $Δ = {(12 x + 196)}^{2} - 84 (91 x^{2} + 406 x + 799) ⩾ 0$ $75 x^{2} + 294 x + 287 ⩽ 0$ $\frac{- 147 - 2 \sqrt{21}}{75} ⩽ x ⩽ \frac{- 147 + 2 \sqrt{21}}{75}$ $\Rightarrow x_{C} = - \frac{147}{75}$ $x = - \frac{147}{75} + ρ \cos θ$ $y = \frac{154}{75} + ρ \sin θ$ $91 {(- \frac{147}{75} + ρ \cos θ)}^{2} + 84 {(\frac{154}{75} + ρ \sin θ)}^{2} - 24 (- \frac{147}{75} + ρ \cos θ) (\frac{154}{75} + ρ \sin θ) + 406 (- \frac{147}{75} + ρ \cos θ) - 392 (\frac{154}{75} + ρ \sin θ) + 799 = 0$ $(175 + 7 \cos 2 θ - 24 \sin 2 θ) ρ^{2} - \frac{8}{3} = 0$ $ρ^{2} = \frac{8}{3 (175 + 7 \cos 2 θ - 24 \sin 2 θ)}$ $ρ^{2} = \frac{8}{3 [175 + 25 (\frac{7}{25} \cos 2 θ - \frac{24}{25} \sin 2 θ)]}$ $ρ^{2} = \frac{8}{75 [7 - \sin (2 θ - \tan^{- 1} \frac{7}{24})]}$ $ρ_{m a x}^{2} = \frac{8}{75 (7 - 1)}$ $\Rightarrow ρ_{m a x} = \frac{2}{15} = s e m i m a j o r a x i s a$ $a t 2 θ - \tan^{- 1} \frac{7}{24} = \frac{π}{2} o r θ_{1} = \frac{π}{4} + \frac{1}{2} \tan^{- 1} \frac{7}{24}$ $ρ_{m i n}^{2} = \frac{8}{75 (7 + 1)}$ $\Rightarrow ρ_{m i n} = \frac{2 \sqrt{6}}{45} = s e m i m i n o r a x i s b$ $a t 2 θ - \tan^{- 1} \frac{7}{24} = - \frac{π}{2} o r θ_{2} = - \frac{π}{4} + \frac{1}{2} \tan^{- 1} \frac{7}{24}$ $\tan θ_{1} = \frac{4}{3}$ $\tan θ_{2} = - \frac{3}{4}$ $. . .$
	Commented byTawa11 last updated on 10/Apr/22

	$Great sir .$
	Commented bymr W last updated on 10/Apr/22

	Commented bymr W last updated on 10/Apr/22

	$a = \frac{2}{15}, b = \frac{2 \sqrt{6}}{45}$ $c = \sqrt{a^{2} - b^{2}} = \frac{2 \sqrt{3}}{45}$ $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \frac{\sqrt{6}}{3}$
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