Question Number 48121 by JDlix last updated on 19/Nov/18 $${can}\:{the}\:{directrix}\:{of}\:{a}\:{parabola}\:{be}\:{in}\:{the}\:{form}\:{y}={mx}+{b}\:\:? \\ $$$${or}\:{is}\:{there}\:{an}\:{inclined}\:{parabola}\:{with}\:{directrix}\:{and}\:{axis}\: \\ $$$${of}\:{symmetry}\:{in}\:{the}\:{form}\:{of}\:{y}={mx}+{b}\:\:?? \\ $$ Commented by MJS last updated on 19/Nov/18 $$\mathrm{you}\:\mathrm{can}\:\mathrm{rotate}\:\mathrm{a}\:\mathrm{parabola},\:\mathrm{so}\:\mathrm{this}\:\mathrm{is}\:\mathrm{indeed} \\…
Question Number 48113 by ajfour last updated on 19/Nov/18 Answered by MJS last updated on 20/Nov/18 $$\mathrm{circle}\:\mathrm{1} \\ $$$${c}_{\mathrm{1}} :\:\left({x}+{o}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} −{o}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{circle}\:\mathrm{2}…
Question Number 113644 by AbhishekBasnet last updated on 14/Sep/20 Answered by 1549442205PVT last updated on 14/Sep/20 $$\mathrm{The}\:\mathrm{intersection}\:\mathrm{point}\:\mathrm{of}\:\mathrm{two}\:\mathrm{lines} \\ $$$$\mathrm{x}−\mathrm{y}+\mathrm{1}=\mathrm{0}\:\mathrm{and}\:\mathrm{2x}−\mathrm{y}−\mathrm{1}=\mathrm{0}\:\mathrm{is}\:\mathrm{roots} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{system}:\begin{cases}{\mathrm{x}−\mathrm{y}+\mathrm{1}=\mathrm{0}}\\{\mathrm{2x}−\mathrm{y}−\mathrm{1}=\mathrm{0}}\end{cases} \\ $$$$\mathrm{Substracting}\:\mathrm{two}\:\:\mathrm{above}\:\mathrm{equations} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{x}−\mathrm{2}=\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{2},\mathrm{y}=\mathrm{3}.\mathrm{Replace}\:…
Question Number 47939 by rahul 19 last updated on 17/Nov/18 Commented by rahul 19 last updated on 17/Nov/18 $${how}\:{to}\:{draw}\:{this}\:{graph}? \\ $$$${steps}\:{plss} \\ $$ Commented by…
Question Number 178833 by Spillover last updated on 22/Oct/22 Answered by cortano1 last updated on 22/Oct/22 $$\left(\mathrm{b}\right)\:\mathrm{L}_{\mathrm{3}} \equiv\:\mathrm{L}_{\mathrm{2}} +\mathrm{k}\left(\mathrm{L}_{\mathrm{1}} −\mathrm{L}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{10x}−\mathrm{12y}+\mathrm{40}+\mathrm{k}\left(\mathrm{8x}+\mathrm{8y}−\mathrm{44}\right)=\mathrm{0}…
Question Number 113262 by bemath last updated on 12/Sep/20 $$\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}} \\…