Question Number 28359 by ajfour last updated on 24/Jan/18 Commented by ajfour last updated on 24/Jan/18 $${Or}\:{rather}\:{the}\:{locus}\:{of}\:{the}\:{centre}\: \\ $$$${of}\:{ellipse}. \\ $$ Terms of Service Privacy…
Question Number 93855 by Ar Brandon last updated on 15/May/20 $$\mathrm{At}\:\mathrm{what}\:\mathrm{point}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parabola}\:\mathrm{y}=\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{will}\:\mathrm{the}\:\mathrm{tangent}\: \\ $$$$\mathrm{a}\backslash\:\mathrm{be}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{line}\:\mathrm{y}=\mathrm{4x}−\mathrm{5} \\ $$$$\mathrm{b}\backslash\:\mathrm{be}\:\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{2x}−\mathrm{6y}+\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{c}\backslash\:\mathrm{make}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{45}^{°} \:\mathrm{with}\:\mathrm{3x}−\mathrm{y}+\mathrm{1}=\mathrm{0} \\ $$ Answered by Kunal12588…
Question Number 93853 by Ar Brandon last updated on 15/May/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{at}\:\mathrm{which}\:\mathrm{the}\:\mathrm{parabola} \\ $$$$\mathrm{y}=\mathrm{x}^{\mathrm{2}} \:\mathrm{cuts}\:\mathrm{through}\:\mathrm{the}\:\mathrm{line}\:\mathrm{3x}−\mathrm{y}−\mathrm{2}=\mathrm{0} \\ $$ Answered by Kunal12588 last updated on 15/May/20 $${y}={x}^{\mathrm{2}} ,\:{y}=\mathrm{3}{x}−\mathrm{2}…
Question Number 28303 by ajfour last updated on 24/Jan/18 $${If}\:{one}\:{line}\:{of}\:{the}\:{equation}\:: \\ $$$$\boldsymbol{{ax}}^{\mathrm{3}} +\boldsymbol{{bx}}^{\mathrm{2}} \boldsymbol{{y}}+\boldsymbol{{cxy}}^{\mathrm{2}} +\boldsymbol{{dy}}^{\mathrm{3}} =\mathrm{0} \\ $$$${bisects}\:{the}\:{angle}\:{between}\:{the} \\ $$$${the}\:{other}\:{two}\:{then}\:{prove} \\ $$$$\left(\mathrm{3}{a}+{c}\right)^{\mathrm{2}} \left({bc}+\mathrm{2}{cd}−\mathrm{3}{ad}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left({b}+\mathrm{3}{d}\right)^{\mathrm{2}}…
Question Number 28278 by ajfour last updated on 23/Jan/18 Commented by ajfour last updated on 23/Jan/18 $${If}\:{y}={mx}+{c}\:\:\:{be}\:{tangent}\:{to}\:{the} \\ $$$${ellipse}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:,\:{find}\:{the} \\ $$$${condition}\:{for}\:{tangency}\:{of}\:{the}\:…
Question Number 28241 by ajfour last updated on 22/Jan/18 Commented by ajfour last updated on 22/Jan/18 $${Q}.\:\mathrm{28185}\:\:\:\left({alternate}\:{solution}\:\right) \\ $$ Answered by ajfour last updated on…
Question Number 28185 by ajfour last updated on 21/Jan/18 $${M},\:{N}\:{are}\:{endpoints}\:{of}\:{a}\:{diameter} \\ $$$$\:\mathrm{4}{x}−{y}=\mathrm{15}\:\:{of}\:{circle} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{6}{y}−\mathrm{16}=\mathrm{0}\:;\:{and}\:{are} \\ $$$${also}\:{on}\:{the}\:{tangents}\:{at}\:{the}\:{end} \\ $$$${points}\:{of}\:{the}\:{major}\:{axis}\:{of}\:{an} \\ $$$${ellipse}\:{respectively},\:{such}\:{that} \\ $$$${MN}\:{is}\:{also}\:{tangent}\:{to}\:{the}\:{same} \\…
Question Number 159180 by cortano last updated on 14/Nov/21 Answered by som(math1967) last updated on 14/Nov/21 $${HI}=\mathrm{50}{tan}\mathrm{48}=\mathrm{55}.\mathrm{53} \\ $$$${HL}=\mathrm{55}.\mathrm{53}+\mathrm{40}=\mathrm{95}.\mathrm{53} \\ $$$${LQ}=\mathrm{50}+\mathrm{8}=\mathrm{58} \\ $$$${HQ}=\sqrt{\mathrm{58}^{\mathrm{2}} +\left(\mathrm{95}.\mathrm{53}\right)^{\mathrm{2}} }=\mathrm{111}.\mathrm{75}\:{feet}\left({approx}\right)…
Question Number 159135 by cherokeesay last updated on 13/Nov/21 Commented by Rasheed.Sindhi last updated on 13/Nov/21 $${I}'{ve}\:{now}\:{understood}\:{completely}\: \\ $$$${although}\:\:{difficultly}. \\ $$$${I}\:{want}\:{to}\:{say}\:{only}\:{that}\:{the}\:{answer} \\ $$$${should}\:{be}\:{somewhat}\:{more}\:{detailed} \\ $$$${in}\:{order}\:{that}\:{the}\:{persons}\:{who}\:{are}…
Question Number 159118 by physicstutes last updated on 13/Nov/21 $$\mathrm{Let}\:{F}\left({r}\right)\:=\:\begin{cases}{{mkr},\:\:\mathrm{0}\:\leqslant\:{r}\:<\:{R}}\\{\frac{{m}\mathrm{g}{R}^{\mathrm{2}} }{{r}^{\mathrm{2}} },\:{r}\geqslant\:{R}}\end{cases} \\ $$$$\mathrm{find}\:{D}_{{F}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com