Question Number 45654 by ajfour last updated on 15/Oct/18 Commented by ajfour last updated on 15/Oct/18 $${Estimate}\:{minimum},\:{maximum}\:\boldsymbol{\phi}\: \\ $$$${in}\:{terms}\:{of}\:{r}\:{and}\:{R}.\:{Equation}\:{of} \\ $$$${ellipse}\:{being}\:\:\:\frac{{x}^{\mathrm{2}} }{{R}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{r}^{\mathrm{2}} }=\mathrm{1}\:.…
Question Number 45638 by peter frank last updated on 14/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18 $$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\:\:\left({asec}\theta,{btan}\theta\right)\:{lies}\:{on}\:{hyperbola} \\ $$$${eqn}\:{of}\:{tangent}\:\:{at}\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}}…
Question Number 45608 by peter frank last updated on 14/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18 $${p}\left(\alpha,\beta\right)\:\:{Q}\left({a},{b}\right)\:\:\:{pointR}\:{devides}\:{PQ}\:\mathrm{2}:\mathrm{1} \\ $$$${R}\left(\frac{\mathrm{2}{a}+\alpha}{\mathrm{3}},\frac{\mathrm{2}{b}+\beta}{\mathrm{3}}\right)=\left({h},{k}\right)\:{say} \\ $$$$\frac{\mathrm{2}{a}+\alpha}{\mathrm{3}}={h}\:\:\:\alpha=\mathrm{3}{h}−\mathrm{2}{a}\:\:\:\: \\ $$$$\frac{\mathrm{2}{b}+\beta}{\mathrm{3}}={k}\:\:\:\:\:\beta=\mathrm{3}{k}−\mathrm{2}{b} \\…
Question Number 45602 by peter frank last updated on 14/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18 $$\left({ct},\frac{{c}}{{t}}\right)\:{lies}\:{on}\:{rectangulsr}\:{hyperbola} \\ $$$${xy}={c}^{\mathrm{2}} \\ $$$$\frac{{d}}{{dx}}\left({xy}\right)=\frac{{d}}{{dx}}\left({c}^{\mathrm{2}} \right) \\ $$$${x}\frac{{dy}}{{dx}}+{y}=\mathrm{0}\:\:\:\frac{{dy}}{{dx}}=−\frac{{y}}{{x}}…
Question Number 45565 by ajfour last updated on 14/Oct/18 $${If}\:\:\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{by}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{hxy}}+\mathrm{2}\boldsymbol{{gx}}+\mathrm{2}\boldsymbol{{fy}}+\boldsymbol{{c}}=\mathrm{0} \\ $$$${be}\:{the}\:{equation}\:{of}\:{an}\:{ellipse},\:{find} \\ $$$${coordinates}\:{of}\:{center}\:{of}\:{ellipse}. \\ $$$${Q}.\mathrm{45506}\:\:\left({another}\:{solution}\right) \\ $$ Answered by ajfour last updated…
Question Number 45514 by peter frank last updated on 13/Oct/18 Commented by peter frank last updated on 14/Oct/18 $$\mathrm{please}\:\mathrm{help}. \\ $$ Terms of Service Privacy…
Question Number 45506 by ajfour last updated on 13/Oct/18 $${If}\:\:\:{ax}^{\mathrm{2}} +{by}^{\mathrm{2}} +\mathrm{2}{hxy}+\mathrm{2}{gx}+\mathrm{2}{fy}+{c}=\mathrm{0} \\ $$$${be}\:{the}\:{equation}\:{of}\:{an}\:{ellipse},\:{find} \\ $$$${coordinates}\:{of}\:{its}\:{centre}. \\ $$ Answered by MrW3 last updated on 14/Oct/18…
Question Number 45477 by ajfour last updated on 13/Oct/18 Commented by ajfour last updated on 13/Oct/18 $${If}\:{radius}\:{of}\:{circle}\:{is}\:{unity}, \\ $$$${find}\:{equation}\:{of}\:{ellipse}. \\ $$$$\left({circle}\:{may}\:{or}\:{may}\:{not}\:{touch}\:{y}-{axes}\right) \\ $$ Commented by…
Question Number 110939 by ajfour last updated on 31/Aug/20 Commented by ajfour last updated on 31/Aug/20 $${If}\:{both}\:{ellipses}\:{are}\:{congruent},\:{find} \\ $$$$\mathrm{tan}\:\alpha. \\ $$ Answered by ajfour last…
Question Number 45208 by Necxx last updated on 10/Oct/18 Commented by Necxx last updated on 10/Oct/18 $${pls}\:{help}\:{with}\:\mathrm{61}\:{qnd}\:\mathrm{62} \\ $$ Terms of Service Privacy Policy Contact:…