Question Number 48156 by ajfour last updated on 20/Nov/18 Commented by ajfour last updated on 20/Nov/18 $${Find}\:{R}\:{in}\:{terms}\:{of}\:{a}. \\ $$ Answered by ajfour last updated on…
Question Number 48143 by rahul 19 last updated on 20/Nov/18 $${The}\:{locus}\:{of}\:{P}\left({x},{y}\right)\:{such}\:{that} \\ $$$$\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{8}{y}+\mathrm{16}}−\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}}=\mathrm{5}\:{is}? \\ $$ Commented by rahul 19 last updated…
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}}…