Question Number 25199 by tawa tawa last updated on 06/Dec/17 $$\mathrm{A}\:\mathrm{line}\:\mathrm{has}\:\mathrm{equation}\:\mathrm{y}\:=\:\mathrm{2x}\:−\:\mathrm{7}\:\mathrm{and}\:\mathrm{a}\:\mathrm{curve}\:\mathrm{has}\:\mathrm{equation}\:\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{4x}\:+\:\mathrm{c}, \\ $$$$\mathrm{where}\:\mathrm{c}\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\:\mathrm{c}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{line} \\ $$$$\mathrm{does}\:\mathrm{not}\:\mathrm{intersect}\:\mathrm{the}\:\mathrm{curve}. \\ $$ Commented by tawa tawa last updated on…
Question Number 25156 by tawa tawa last updated on 05/Dec/17 Answered by nnnavendu last updated on 05/Dec/17 $${we}\:{know}\:{that}\:{point}\:{on}\:{the}\:{x}\:−\:{asix}\:{A}\equiv\left(−{x},\mathrm{0}\right) \\ $$$${then}\:{from}\:{distance}\:{formula}\:{and}\:{diagram} \\ $$$${AB}={AD} \\ $$$$\sqrt{\left(−{x}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{0}−\mathrm{10}\right)^{\mathrm{2}}…
Question Number 25002 by ajfour last updated on 30/Nov/17 Commented by ajfour last updated on 30/Nov/17 $${Solution}\:{to}\:{Q}.\mathrm{24966} \\ $$ Answered by ajfour last updated on…
Question Number 90520 by ajfour last updated on 24/Apr/20 Commented by ajfour last updated on 24/Apr/20 $${If}\:{eq}.\:{of}\:{ellipse}\:{is}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:, \\ $$$${find}\:{equation}\:{of}\:{shown}\:{parabola}. \\ $$…
Question Number 24966 by math solver last updated on 30/Nov/17 $$\mathrm{find}\:\mathrm{the}\:\mathrm{focus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hyperbola}\: \\ $$$${x}^{\mathrm{2}} −\mathrm{16}{xy}−\mathrm{11}{y}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{6}{y}+\mathrm{21}=\mathrm{0}\:? \\ $$ Commented by math solver last updated on 30/Nov/17…
Question Number 90499 by ajfour last updated on 24/Apr/20 Commented by ajfour last updated on 24/Apr/20 $$\bigstar{Similar}\:{to}\:{a}\:{previous}\:{question}. \\ $$$${If}\:{parabola}\:{is}\:{y}={x}^{\mathrm{2}} ,\:{and}\:\bigtriangleup{ABC} \\ $$$${is}\:{equilateral},\:{find}\:{eq}.\:{of}\:{circle}. \\ $$ Commented…
Question Number 90437 by ajfour last updated on 23/Apr/20 Commented by ajfour last updated on 23/Apr/20 $${If}\:{AB}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:,\:{find}\:{max}.\:{and} \\ $$$${min}.\:{area}\:{of}\:\bigtriangleup{ABC}. \\ $$ Commented by…
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Question Number 90326 by ajfour last updated on 22/Apr/20 Commented by ajfour last updated on 22/Apr/20 $${Find}\:{a}/{b}.\:{Ellipse}\:\:{b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} {y}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} . \\ $$…
Question Number 155866 by cherokeesay last updated on 05/Oct/21 Answered by talminator2856791 last updated on 05/Oct/21 $$\:\mathrm{2}\sqrt{\mathrm{3}}\:+\:\mathrm{2}\sqrt{\mathrm{3}}\:−\:\frac{\mathrm{9}\pi\centerdot\mathrm{arctan}\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}{\mathrm{2}\pi}\:−\:\frac{\pi}{\mathrm{4}}\:−\:\frac{\mathrm{arctan}\left(\frac{\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)}{\mathrm{2}}\:\: \\ $$$$\:=\:\mathrm{4}\sqrt{\mathrm{3}}\:−\:\frac{\mathrm{9}}{\mathrm{2}}\:\mathrm{arctan}\left(\sqrt{\mathrm{3}}\right)−\:\frac{\pi}{\mathrm{4}}\:−\:\mathrm{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\:=\:\mathrm{4}\sqrt{\mathrm{3}}\:−\:\frac{\mathrm{9}\pi}{\mathrm{6}}\:−\:\frac{\pi}{\mathrm{4}}\:−\:\frac{\pi}{\mathrm{6}} \\ $$$$\:=\:\mathrm{4}\sqrt{\mathrm{3}}\:−\:\frac{\mathrm{13}\pi}{\mathrm{12}} \\ $$…