Question Number 113262 by bemath last updated on 12/Sep/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{form}\:\mathrm{of}\: \\ $$$$\mathrm{equation}\:\mathrm{4x}^{\mathrm{2}} −\mathrm{2y}^{\mathrm{2}} =\mathrm{5} \\ $$ Answered by mr W last updated on 12/Sep/20 $$\frac{{x}^{\mathrm{2}}…
Question Number 47621 by rahul 19 last updated on 12/Nov/18 $${Find}\:{locus}\:{of}\:{point}\:{P}\:{from}\:{which} \\ $$$${tangents}\:{PA}\:\&\:{PB}\:{to}\:{circles}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${and}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={b}^{\mathrm{2}} \:{respectively}\:{are}\:{perpendicular}. \\ $$ Commented by rahul…
Question Number 178647 by infinityaction last updated on 19/Oct/22 Answered by mr W last updated on 19/Oct/22 $${side}\:{length}\:={x} \\ $$$$\frac{{x}}{\mathrm{5}}=\frac{\mathrm{5}+\mathrm{1}}{{x}+\sqrt{\mathrm{5}^{\mathrm{2}} −{x}^{\mathrm{2}} }} \\ $$$${x}\sqrt{\mathrm{25}−{x}^{\mathrm{2}} }=\mathrm{30}−{x}^{\mathrm{2}}…
Question Number 113080 by bobhans last updated on 11/Sep/20 $$\mathrm{The}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{is} \\ $$$$\mathrm{84}\:\mathrm{cm}\:\mathrm{and}\:\mathrm{it}'\mathrm{s}\:\mathrm{area}\:\mathrm{is}\:\mathrm{336}\:\mathrm{square}\:\mathrm{cm}.\:\mathrm{If}\:\mathrm{the}\: \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{30}\:\mathrm{cm},\:\mathrm{then} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{two} \\ $$$$\mathrm{sides}\:\mathrm{of}\:\mathrm{triangle}\:? \\ $$ Answered by bemath last updated…
Question Number 113059 by bemath last updated on 11/Sep/20 Answered by john santu last updated on 11/Sep/20 $$\:\frac{\mathrm{1}}{\mathrm{2}.\mathrm{3}.\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{3}.\mathrm{4}.\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{5}.\mathrm{6}.\mathrm{7}}\:= \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right).\left({k}+\mathrm{2}\right).\left({k}+\mathrm{3}\right)}\: \\ $$$$\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)\left({k}+\mathrm{3}\right)}=\frac{{p}}{{k}+\mathrm{1}}+\frac{{q}}{{k}+\mathrm{2}}+\frac{{r}}{{k}+\mathrm{3}} \\…
Question Number 47513 by ajfour last updated on 11/Nov/18 Commented by ajfour last updated on 11/Nov/18 $${Find}\:{coordinates}\:{of}\:{all}\:{points} \\ $$$${that}\:{have}\:{same}\:\bot\:{distance}\:{from} \\ $$$${from}\:{the}\:{three}\:{lines}. \\ $$$${For}\:{each}\:{such}\:{point}\:{this}\:\bot \\ $$$${distance}\:{may}\:{differ}.…
Question Number 47480 by ajfour last updated on 10/Nov/18 Answered by MrW3 last updated on 10/Nov/18 $${assume}\:{the}\:{major}\:{and}\:{minor}\:{axes} \\ $$$${are}\:{parallel}\:{to}\:{x}\:{axis}\:{and}\:{y}\:{axis}. \\ $$$${eqn}. \\ $$$$\frac{\left({x}−{h}\right)^{\mathrm{2}} }{{u}^{\mathrm{2}} }+\frac{\left({y}−{k}\right)^{\mathrm{2}}…
Question Number 47457 by rahul 19 last updated on 10/Nov/18 $${A}\:{straight}\:{line}\:{through}\:\left(\mathrm{2},\mathrm{2}\right)\:{intersects} \\ $$$${lines}\:\sqrt{\mathrm{3}}{x}+{y}=\mathrm{0}\:{and}\:\sqrt{\mathrm{3}}{x}−{y}=\mathrm{0}\:{at}\:{pts}. \\ $$$${A}\:\&\:{B}\:{respectively}.\:{Find}\:{equation} \\ $$$${of}\:{line}\:{AB}\:{so}\:{that}\:\Delta{OAB}\:{is}\:{equilateral}? \\ $$ Answered by rahul 19 last updated…
Question Number 47455 by ajfour last updated on 10/Nov/18 Commented by ajfour last updated on 10/Nov/18 $${Find}\:{maximum}\:{area}\:{of}\:{inscribed} \\ $$$${ellipse}\:{within}\:{AP},\:{BP},\:\&\:{x},{y}\:{axes}. \\ $$ Terms of Service Privacy…
Question Number 47438 by rahul 19 last updated on 10/Nov/18 $${Find}\:{locus}\:{of}\:{a}\:{point}\:{P}\:{which}\:{moves} \\ $$$${such}\:{that}\:{its}\:{distance}\:{from}\:{the}\:{line} \\ $$$${y}=\sqrt{\mathrm{3}}{x}−\mathrm{7}\:{is}\:{same}\:{as}\:{its}\:{distance}\:{from} \\ $$$$\left(\mathrm{2}\sqrt{\mathrm{3}},−\mathrm{1}\right)\:? \\ $$ Answered by ajfour last updated on…