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Question Number 31846 by momo last updated on 15/Mar/18 $$\mathrm{25}\left[\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\mathrm{3}\right)^{\mathrm{2}} \right]=\left(\mathrm{3}{x}−\mathrm{4}{y}+\mathrm{7}\right)^{\mathrm{2}} \\ $$$${is}\:{the}\:{equation}\:{of}\:{parabola}.{Find} \\ $$$${length}\:{of}\:{latus}\:{rectum} \\ $$ Commented by momo last updated on 16/Mar/18…
Question Number 162860 by Mathematification last updated on 01/Jan/22 Answered by MJS_new last updated on 01/Jan/22 $${f}\left({x}\right)={x}^{\mathrm{4}} −\mathrm{18}{x}^{\mathrm{2}} −{x}+\mathrm{83} \\ $$$${g}\left({x}\right)={c}_{\mathrm{1}} {x}+{c}_{\mathrm{0}} \\ $$$${h}\left({x}\right)={f}\left({x}\right)−{g}\left({x}\right)={x}^{\mathrm{4}} −\mathrm{18}{x}^{\mathrm{2}}…
Question Number 31639 by Tinkutara last updated on 11/Mar/18 Answered by MJS last updated on 11/Mar/18 $${P}\in{par}:\:\begin{pmatrix}{{p}}\\{\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−\frac{{p}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{4}}}\end{pmatrix} \\ $$$${y}'\left({p}\right)=\frac{{p}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\:\left(={k}\:\mathrm{of}\:\mathrm{tangent}\:\mathrm{in}\:\mathrm{P}\right) \\ $$$${k}\:\mathrm{of}\:\mathrm{normal}\:{n}\:\mathrm{in}\:\mathrm{P}:\:−\left(\frac{{p}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{1}} = \\ $$$$=−\frac{\mathrm{2}}{{p}−\mathrm{1}}…
Question Number 97143 by Ar Brandon last updated on 06/Jun/20 $$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}}{\mathrm{x}}\:\mathrm{and}\:\mathcal{C}_{\mathrm{f}} \:\mathrm{its}\:\mathrm{graph}; \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{a},\:\mathrm{b},\:\mathrm{and}\:\mathrm{c}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathcal{C}_{\mathrm{f}} \:\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{points}\:\mathrm{A}\left(\mathrm{1};\mathrm{2}\right);\:\mathrm{B}\left(−\mathrm{4};\mathrm{8}\right)\:\mathrm{and}\:\mathrm{has} \\ $$$$\mathrm{a}\:\mathrm{tangent}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{x}−\mathrm{axis}\:\mathrm{at}\:\mathrm{x}=\mathrm{2}. \\ $$ Answered by MJS…
Question Number 162651 by Mathematification last updated on 31/Dec/21 Answered by tounghoungko last updated on 31/Dec/21 $$\:{y}=−{x}\left({x}+{a}\right)^{\mathrm{2}} \:=−{x}^{\mathrm{3}} −\mathrm{2}{ax}^{\mathrm{2}} −{a}^{\mathrm{2}} {x} \\ $$$$\:{y}'=−\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{ax}−{a}^{\mathrm{2}} \:=\:\mathrm{0}…
Question Number 162561 by Mathematification last updated on 30/Dec/21 Answered by MJS_new last updated on 30/Dec/21 $$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mid{AB}\mid \\ $$$$\mathrm{with}\:{A}=\begin{pmatrix}{{p}}\\{{p}^{\mathrm{2}} }\end{pmatrix}\:\mathrm{and}\:{B}=\begin{pmatrix}{{q}}\\{\mathrm{ln}\:{q}}\end{pmatrix}\:.\:\mathrm{this}\:\mathrm{can}\:\mathrm{only} \\ $$$$\mathrm{be}\:\mathrm{approximated}.\:\mathrm{I}\:\mathrm{get} \\ $$$${p}\approx.\mathrm{538167} \\…
Question Number 162496 by cortano last updated on 29/Dec/21 Answered by mr W last updated on 30/Dec/21 Commented by mr W last updated on 30/Dec/21…
Question Number 31406 by rahul 19 last updated on 08/Mar/18 Commented by mrW2 last updated on 08/Mar/18 $${OM}_{{r}} =\frac{{OA}_{{r}} +{OA}_{{r}+\mathrm{1}} }{\mathrm{2}}=\frac{\mathrm{1}+{q}}{\mathrm{2}}×{OA}_{{r}} \\ $$$$\Sigma{OM}_{{r}} =\frac{\mathrm{1}+{q}}{\mathrm{2}}×\Sigma{OA}_{{r}} =\frac{\mathrm{1}+{q}}{\mathrm{2}\left(\mathrm{1}−{q}\right)}×{OA}_{\mathrm{1}}…
Question Number 31383 by rahul 19 last updated on 07/Mar/18 Commented by rahul 19 last updated on 07/Mar/18 $${but}\:{this}\:{ans}.\:{i}\:{am}\:{getting}\:{by}\:{assuming} \\ $$$${isosceles}\:{triangle}.\:{for}\:{general}\:{triangle} \\ $$$${how}\:{can}\:{do}\:? \\ $$…