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

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

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

Answered by mr W last updated on 10/Apr/22

91x^2 −2(12y−203)x+84y^2 −392y+799=0  Δ=(12y−203)^2 −91(84y^2 −392y+799)≥0  75y^2 −308y+315≤0  ((154−(√(91)))/(75))≤y≤((154+(√(91)))/(75))  ⇒y_C =((154)/(75))  84y^2 −2(12x+196)y+91x^2 +406x+799=0  Δ=(12x+196)^2 −84(91x^2 +406x+799)≥0  75x^2 +294x+287≤0  ((−147−2(√(21)))/(75))≤x≤((−147+2(√(21)))/(75))  ⇒x_C =−((147)/(75))  x=−((147)/(75))+ρ cos θ  y=((154)/(75))+ρ sin θ  91(−((147)/(75))+ρ cos θ)^2 +84(((154)/(75))+ρ sin θ)^2 −24(−((147)/(75))+ρ cos θ)(((154)/(75))+ρ sin θ)+406(−((147)/(75))+ρ cos θ)−392(((154)/(75))+ρ sin θ)+799=0  (175+7 cos 2θ−24 sin 2θ)ρ^2 −(8/3)=0  ρ^2 =(8/(3(175+7 cos 2θ−24 sin 2θ)))  ρ^2 =(8/(3[175+25((7/(25)) cos 2θ−((24)/(25)) sin 2θ)]))  ρ^2 =(8/(75[7−sin (2θ−tan^(−1) (7/(24)))]))  ρ_(max) ^2 =(8/(75(7−1)))  ⇒ρ_(max) =(2/(15))=semi major axis a  at 2θ−tan^(−1) (7/(24))=(π/2) or θ_1 =(π/4)+(1/2)tan^(−1) (7/(24))  ρ_(min) ^2 =(8/(75(7+1)))  ⇒ρ_(min) =((2(√6))/(45))=semi minor axis b   at 2θ−tan^(−1) (7/(24))=−(π/2) or θ_2 =−(π/4)+(1/2)tan^(−1) (7/(24))  tan θ_1 =(4/3)  tan θ_2 =−(3/4)  ...

$$\mathrm{91}{x}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{12}{y}−\mathrm{203}\right){x}+\mathrm{84}{y}^{\mathrm{2}} −\mathrm{392}{y}+\mathrm{799}=\mathrm{0} \\ $$ $$\Delta=\left(\mathrm{12}{y}−\mathrm{203}\right)^{\mathrm{2}} −\mathrm{91}\left(\mathrm{84}{y}^{\mathrm{2}} −\mathrm{392}{y}+\mathrm{799}\right)\geqslant\mathrm{0} \\ $$ $$\mathrm{75}{y}^{\mathrm{2}} −\mathrm{308}{y}+\mathrm{315}\leqslant\mathrm{0} \\ $$ $$\frac{\mathrm{154}−\sqrt{\mathrm{91}}}{\mathrm{75}}\leqslant{y}\leqslant\frac{\mathrm{154}+\sqrt{\mathrm{91}}}{\mathrm{75}} \\ $$ $$\Rightarrow{y}_{{C}} =\frac{\mathrm{154}}{\mathrm{75}} \\ $$ $$\mathrm{84}{y}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{12}{x}+\mathrm{196}\right){y}+\mathrm{91}{x}^{\mathrm{2}} +\mathrm{406}{x}+\mathrm{799}=\mathrm{0} \\ $$ $$\Delta=\left(\mathrm{12}{x}+\mathrm{196}\right)^{\mathrm{2}} −\mathrm{84}\left(\mathrm{91}{x}^{\mathrm{2}} +\mathrm{406}{x}+\mathrm{799}\right)\geqslant\mathrm{0} \\ $$ $$\mathrm{75}{x}^{\mathrm{2}} +\mathrm{294}{x}+\mathrm{287}\leqslant\mathrm{0} \\ $$ $$\frac{−\mathrm{147}−\mathrm{2}\sqrt{\mathrm{21}}}{\mathrm{75}}\leqslant{x}\leqslant\frac{−\mathrm{147}+\mathrm{2}\sqrt{\mathrm{21}}}{\mathrm{75}} \\ $$ $$\Rightarrow{x}_{{C}} =−\frac{\mathrm{147}}{\mathrm{75}} \\ $$ $${x}=−\frac{\mathrm{147}}{\mathrm{75}}+\rho\:\mathrm{cos}\:\theta \\ $$ $${y}=\frac{\mathrm{154}}{\mathrm{75}}+\rho\:\mathrm{sin}\:\theta \\ $$ $$\mathrm{91}\left(−\frac{\mathrm{147}}{\mathrm{75}}+\rho\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\mathrm{84}\left(\frac{\mathrm{154}}{\mathrm{75}}+\rho\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} −\mathrm{24}\left(−\frac{\mathrm{147}}{\mathrm{75}}+\rho\:\mathrm{cos}\:\theta\right)\left(\frac{\mathrm{154}}{\mathrm{75}}+\rho\:\mathrm{sin}\:\theta\right)+\mathrm{406}\left(−\frac{\mathrm{147}}{\mathrm{75}}+\rho\:\mathrm{cos}\:\theta\right)−\mathrm{392}\left(\frac{\mathrm{154}}{\mathrm{75}}+\rho\:\mathrm{sin}\:\theta\right)+\mathrm{799}=\mathrm{0} \\ $$ $$\left(\mathrm{175}+\mathrm{7}\:\mathrm{cos}\:\mathrm{2}\theta−\mathrm{24}\:\mathrm{sin}\:\mathrm{2}\theta\right)\rho^{\mathrm{2}} −\frac{\mathrm{8}}{\mathrm{3}}=\mathrm{0} \\ $$ $$\rho^{\mathrm{2}} =\frac{\mathrm{8}}{\mathrm{3}\left(\mathrm{175}+\mathrm{7}\:\mathrm{cos}\:\mathrm{2}\theta−\mathrm{24}\:\mathrm{sin}\:\mathrm{2}\theta\right)} \\ $$ $$\rho^{\mathrm{2}} =\frac{\mathrm{8}}{\mathrm{3}\left[\mathrm{175}+\mathrm{25}\left(\frac{\mathrm{7}}{\mathrm{25}}\:\mathrm{cos}\:\mathrm{2}\theta−\frac{\mathrm{24}}{\mathrm{25}}\:\mathrm{sin}\:\mathrm{2}\theta\right)\right]} \\ $$ $$\rho^{\mathrm{2}} =\frac{\mathrm{8}}{\mathrm{75}\left[\mathrm{7}−\mathrm{sin}\:\left(\mathrm{2}\theta−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{7}}{\mathrm{24}}\right)\right]} \\ $$ $$\rho_{{max}} ^{\mathrm{2}} =\frac{\mathrm{8}}{\mathrm{75}\left(\mathrm{7}−\mathrm{1}\right)} \\ $$ $$\Rightarrow\rho_{{max}} =\frac{\mathrm{2}}{\mathrm{15}}={semi}\:{major}\:{axis}\:{a} \\ $$ $${at}\:\mathrm{2}\theta−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{7}}{\mathrm{24}}=\frac{\pi}{\mathrm{2}}\:{or}\:\theta_{\mathrm{1}} =\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{7}}{\mathrm{24}} \\ $$ $$\rho_{{min}} ^{\mathrm{2}} =\frac{\mathrm{8}}{\mathrm{75}\left(\mathrm{7}+\mathrm{1}\right)} \\ $$ $$\Rightarrow\rho_{{min}} =\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{45}}={semi}\:{minor}\:{axis}\:{b}\: \\ $$ $${at}\:\mathrm{2}\theta−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{7}}{\mathrm{24}}=−\frac{\pi}{\mathrm{2}}\:{or}\:\theta_{\mathrm{2}} =−\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{7}}{\mathrm{24}} \\ $$ $$\mathrm{tan}\:\theta_{\mathrm{1}} =\frac{\mathrm{4}}{\mathrm{3}} \\ $$ $$\mathrm{tan}\:\theta_{\mathrm{2}} =−\frac{\mathrm{3}}{\mathrm{4}} \\ $$ $$... \\ $$

Commented byTawa11 last updated on 10/Apr/22

Great sir.

$$\mathrm{Great}\:\mathrm{sir}. \\ $$

Commented bymr W last updated on 10/Apr/22

Commented bymr W last updated on 10/Apr/22

a=(2/(15)), b=((2(√6))/(45))  c=(√(a^2 −b^2 ))=((2(√3))/(45))  e=(√(1−(b^2 /a^2 )))=((√6)/3)

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

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