Question Number 49740 by ajfour last updated on 09/Dec/18 Commented by ajfour last updated on 09/Dec/18 $${Find}\:{side}\:{length}\:\boldsymbol{{s}}\:{of}\:{a}\:{pentagon} \\ $$$${of}\:{equal}\:{sides}\:{inscribed}\:{within} \\ $$$${the}\:{ellipse},\:{in}\:{terms}\:{of}\:{ellipse} \\ $$$${parameters}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}. \\ $$…
Question Number 49737 by ajfour last updated on 09/Dec/18 Commented by ajfour last updated on 09/Dec/18 $${Find}\:{parameters}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}\:{of} \\ $$$${ellipse}\:{circumscribing}\:{a}\:{rectangle} \\ $$$${of}\:{sides}\:\boldsymbol{{l}}\:{and}\:\boldsymbol{{h}}. \\ $$ Answered by…
Question Number 49526 by ajfour last updated on 07/Dec/18 Commented by ajfour last updated on 07/Dec/18 $${If}\:{both}\:{the}\:{coloured}\:{areas}\:{are}\:{equal}, \\ $$$${find}\:{equation}\:{of}\:{parabola}\:{in}\:{terms} \\ $$$${of}\:{ellipse}\:{parameters}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}. \\ $$ Answered by…
Question Number 115034 by bobhans last updated on 23/Sep/20 $${what}\:{the}\:{equation}\:{of}\:{the}\:{hyperbola}\: \\ $$$${with}\:{the}\:{given}\:{asymtotes}\:{y}=\mathrm{43}{x}+\mathrm{13} \\ $$$${and}\:{y}=−\mathrm{43}{x}+\mathrm{13}\:,\:{a}\:{vertex}\:{at}\:\left(−\mathrm{1},\mathrm{7}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 115035 by bobhans last updated on 23/Sep/20 $${solve}\:\mid\:\mathrm{4}−\frac{\mathrm{3}}{{x}}\:\mid\:<\:\mathrm{8} \\ $$ Commented by bemath last updated on 23/Sep/20 $$\Leftrightarrow\:−\mathrm{8}\:<\:\mathrm{4}−\frac{\mathrm{3}}{{x}}\:<\:\mathrm{8} \\ $$$${case}\left(\mathrm{1}\right)\:\rightarrow\:−\mathrm{8}\:<\:\mathrm{4}−\frac{\mathrm{3}}{{x}} \\ $$$$\:\:\:\:\:\frac{\mathrm{3}}{{x}}\:<\:\mathrm{12}\:;\:\frac{\mathrm{1}}{{x}}\:<\:\mathrm{4}\:;\:\frac{\mathrm{1}−\mathrm{4}{x}}{{x}}\:<\:\mathrm{0} \\…
Question Number 115031 by bobhans last updated on 23/Sep/20 $$\:{circle}\:{of}\:{centre}\:{P}\:\:{touches}\:{externally}\:{both} \\ $$$${the}\:{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}=\mathrm{0}\:{and}\: \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{6}{y}+\mathrm{5}=\mathrm{0}\:.\:{The}\:{locus}\:{of}\:{P}\:{is} \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}{x}^{\mathrm{2}} −\mathrm{3}{xy}+\mathrm{2}{y}^{\mathrm{2}} \:=\:\lambda\left({y}−{x}\right)\:{where}\:\lambda\:{is}\:\_\_ \\ $$ Answered…
Question Number 49468 by munnabhai455111@gmail.com last updated on 07/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18 $${O}\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:\:{A}\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} ,{z}_{\mathrm{1}} \right)\:{B}\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} ,{z}_{\mathrm{2}} \right) \\ $$$${O}\overset{\rightarrow}…
Question Number 49466 by munnabhai455111@gmail.com last updated on 07/Dec/18 $${show}\:{that}\:{the}\:{area}\:{of}\:{the}\:{triangle}\:{whose}\:{vertics}\:{area}\:\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:,\:\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} ,{z}_{\mathrm{1}} \right)\:,\:\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2},} {z}_{\mathrm{2}} \right)\:{is}\:\mathrm{1}/\mathrm{2}\left(\sqrt{\Sigma\left({y}_{\mathrm{1}} {z}_{\mathrm{2}} −{y}_{\mathrm{2}} {z}_{\mathrm{1}} \right)^{\mathrm{2}} \:.}\right. \\ $$ Terms of…
Question Number 49394 by ajfour last updated on 06/Dec/18 Commented by ajfour last updated on 07/Dec/18 $${Find}\:{equation}\:{of}\:{maximum}\:{area} \\ $$$${ellipse}. \\ $$ Answered by ajfour last…
Question Number 49112 by Pk1167156@gmail.com last updated on 03/Dec/18 Commented by Pk1167156@gmail.com last updated on 03/Dec/18 $$\mathrm{find}\:\theta \\ $$ Terms of Service Privacy Policy Contact:…