Question Number 141171 by iloveisrael last updated on 16/May/21 Commented by iloveisrael last updated on 16/May/21 $$\:{insteresting}\:{question} \\ $$ Commented by mr W last updated…
Question Number 75580 by lalitchand last updated on 13/Dec/19 $$\mathrm{Define}\:\mathrm{the}\:\mathrm{inversion}\:\mathrm{transformation} \\ $$$$\mathrm{of}\:\mathrm{Z}=\mathrm{a}\:+\:\mathrm{ib}\:\mathrm{where}\:\mathrm{i}\:\mathrm{is}\:\mathrm{imaginary}\:\mathrm{line}\: \\ $$$$\mathrm{passing}\:\mathrm{through}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{inversion}. \\ $$ Commented by MJS last updated on 13/Dec/19 $$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{clear}.\:\mathrm{i}\:\mathrm{is}\:\mathrm{an}\:\mathrm{imaginary}\:\mathrm{line}? \\…
Question Number 9861 by Raja Naik last updated on 09/Jan/17 $$\mathrm{find}\:\:\lambda\:\mathrm{value}\:\mathrm{in}\:\lambda^{\mathrm{2}} −\mathrm{4}\lambda−\mathrm{13}=\mathrm{0} \\ $$ Answered by FilupSmith last updated on 09/Jan/17 $$\lambda=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}+\mathrm{4}\left(\mathrm{13}\right)}}{\mathrm{2}} \\ $$$$\lambda=\frac{\mathrm{4}\pm\sqrt{\mathrm{68}}}{\mathrm{2}} \\…
Question Number 9731 by tawakalitu last updated on 29/Dec/16 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{asymptotes}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hypebola}\:\mathrm{whose} \\ $$$$\mathrm{equation}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}. \\ $$$$\frac{\mathrm{x}}{\mathrm{24}}\:−\:\frac{\mathrm{y}}{\mathrm{29}}\:=\:\mathrm{1} \\ $$ Commented by geovane10math last updated on 29/Dec/16 $$\frac{\mathrm{29}{x}\:−\:\mathrm{24}{y}}{\mathrm{696}}\:=\:\mathrm{1} \\…
Question Number 9732 by tawakalitu last updated on 29/Dec/16 $$\mathrm{Find}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{an}\:\mathrm{ellipse}\:\mathrm{whose}\:\mathrm{major}\:\mathrm{axis} \\ $$$$\mathrm{is}\:\mathrm{vertical},\:\mathrm{with}\:\mathrm{the}\:\mathrm{center}\:\mathrm{located}\:\left(−\:\mathrm{1},\:\mathrm{3}\right) \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{center}\:\mathrm{and}\:\mathrm{one}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{covertices}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{4},\:\mathrm{and}\:\mathrm{the}\:\mathrm{distance} \\ $$$$\mathrm{between}\:\mathrm{the}\:\mathrm{center}\:\mathrm{and}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vertices}\: \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{6}. \\ $$ Answered by sandy_suhendra…
Question Number 9246 by sandipkd@ last updated on 25/Nov/16 Commented by sandipkd@ last updated on 25/Nov/16 $${question}\:{no}.\:\mathrm{4} \\ $$ Commented by sandipkd@ last updated on…
show-that-the-ellipse-with-e-5-3-focus-0-2-and-directrix-x-4-5-3-has-the-equation-x-5-2-9-y-2-2-4-1-
Question Number 9163 by tawakalitu last updated on 21/Nov/16 $$\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{ellipse}\:\mathrm{with}\:\mathrm{e}\:=\:\frac{\sqrt{\mathrm{5}}}{\mathrm{3}},\: \\ $$$$\mathrm{focus}\:\left(\mathrm{0},\:\mathrm{2}\right)\:\mathrm{and}\:\mathrm{directrix}\:\mathrm{x}\:=\:−\frac{\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$$$\mathrm{has}\:\mathrm{the}\:\mathrm{equation}\::\:\frac{\left(\mathrm{x}\:−\:\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }{\mathrm{9}}\:+\:\frac{\left(\mathrm{y}\:−\:\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}\:=\:\mathrm{1} \\ $$ Commented by sandy_suhendra last updated on 22/Nov/16…
Question Number 74698 by ajfour last updated on 29/Nov/19 Commented by ajfour last updated on 29/Nov/19 $${If}\:\overset{\frown} {{AB}}\:=\overset{\frown} {{AE}}\:=\:\overset{\frown} {{DE}}\:. \\ $$$${Find}\:{equation}\:{of}\:{circle}. \\ $$ Answered…
Question Number 74649 by ajfour last updated on 28/Nov/19 Commented by ajfour last updated on 28/Nov/19 $$\:\:{Find}\:\theta_{{max}} \:{in}\:{terms}\:{of}\:{a},{b},{c}. \\ $$$${The}\:{boundary}\:{is}\:{an}\:{ellipse}. \\ $$ Answered by ajfour…
Question Number 9048 by tawakalitu last updated on 16/Nov/16 Answered by Rasheed Soomro last updated on 16/Nov/16 $$\left.\mathrm{a}\right)\:\:\mathrm{Straight}\:\mathrm{line}:\:\mathrm{y}=\mathrm{mx}+\mathrm{c} \\ $$$$\:\:\:\:\:\:\:\mathrm{Circle}:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{2gx}+\mathrm{2fy}+\mathrm{C}_{\mathrm{1}} =\mathrm{0} \\ $$$$\mathrm{For}\:\mathrm{intersection}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{above}…