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…
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Question Number 44988 by peter frank last updated on 07/Oct/18 $$\boldsymbol{\mathrm{A}}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{ellipse}\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}\:} }+\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }=\mathrm{1}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{point}}\:\boldsymbol{\mathrm{p}}\:\boldsymbol{\mathrm{meets}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{minor}}\:\boldsymbol{\mathrm{axis}} \\ $$$$\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{L}}\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{normal}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{p}}\:\boldsymbol{\mathrm{meets}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{major}}\:\boldsymbol{\mathrm{axis}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{m}}.\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{locus}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{midpoint}}\:\boldsymbol{\mathrm{LM}} \\ $$ Answered by MrW3 last updated on…
Question Number 44891 by peter frank last updated on 06/Oct/18 Answered by MrW3 last updated on 07/Oct/18 $${LineL}:\:{ax}+{by}+{c}=\mathrm{0} \\ $$$${Point}\:{P}_{\mathrm{1}} \:\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right) \\ $$$${Point}\:{P}\left({u},{v}\right)\:{with}\:{PP}_{\mathrm{1}}…