Question Number 111988 by ajfour last updated on 05/Sep/20 Commented by ajfour last updated on 05/Sep/20 $${If}\:\bigtriangleup{ABC}\:{is}\:{equilateral}\:{with}\:{side}\:\boldsymbol{{s}}, \\ $$$${find}\:{x}-{coordinate}\:{of}\:{point}\:{A}\:{in}\: \\ $$$${terms}\:{of}\:{s}. \\ $$ Answered by…
Question Number 177152 by cortano1 last updated on 01/Oct/22 Answered by Ar Brandon last updated on 01/Oct/22 $${x}+{y}=\mathrm{13}+\sqrt{\mathrm{2}}\:…\mathrm{eqn}\left({i}\right) \\ $$$${y}+{z}=\mathrm{20}−\mathrm{3}\sqrt{\mathrm{2}}\:…\mathrm{eqn}\left({ii}\right) \\ $$$${z}+{x}=\mathrm{17}−\mathrm{10}\sqrt{\mathrm{2}}\:…\mathrm{eqn}\left({iii}\right) \\ $$$$\left({i}\right)−\left({ii}\right)+\left({iii}\right) \\…
Question Number 46011 by peter frank last updated on 19/Oct/18 $$\boldsymbol{\mathrm{P}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{any}}\:\boldsymbol{\mathrm{point}}\:\boldsymbol{\mathrm{on}}\:\:\boldsymbol{\mathrm{rectangular}} \\ $$$$\boldsymbol{\mathrm{xy}}=\boldsymbol{\mathrm{c}}^{\mathrm{2}} \boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{line}}\:\boldsymbol{\mathrm{joining}} \\ $$$$\boldsymbol{\mathrm{P}}\:\mathrm{to}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{and}\:\mathrm{the}\: \\ $$$$\mathrm{tangent}\:\mathrm{at}\:\mathrm{P}\:\mathrm{are}\:\mathrm{equally}\:\mathrm{inclined} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{assymptotes} \\ $$ Terms of Service…
Question Number 46012 by peter frank last updated on 19/Oct/18 $$\mathrm{the}\:\mathrm{normal}\:\mathrm{at}\:\mathrm{any}\:\mathrm{point} \\ $$$$\mathrm{of}\:\mathrm{hyperbola}\:\mathrm{meets}\:\mathrm{the}\:\mathrm{axes} \\ $$$$\mathrm{at}\:\mathrm{E},\mathrm{F}.\mathrm{find}\:\mathrm{the}\:\mathrm{locus}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{EF}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on…
Question Number 176811 by cortano1 last updated on 27/Sep/22 Commented by cortano1 last updated on 27/Sep/22 $$\mathrm{i}\:\mathrm{got}\:\mathrm{3}\::\:\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact:…
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…