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Category: Coordinate Geometry

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-

Question Number 48121 by JDlix last updated on 19/Nov/18 canthedirectrixofaparabolabeintheformy=mx+b?oristhereaninclinedparabolawithdirectrixandaxisofsymmetryintheformofy=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-113644

Question Number 113644 by AbhishekBasnet last updated on 14/Sep/20 Answered by 1549442205PVT last updated on 14/Sep/20 Theintersectionpointoftwolinesxy+1=0and2xy1=0isrootsofthesystem:{xy+1=02xy1=0Substractingtwoaboveequations$$\mathrm{we}\:\mathrm{get}\:\mathrm{x}−\mathrm{2}=\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{2},\mathrm{y}=\mathrm{3}.\mathrm{Replace}\:…

Question-113059

Question Number 113059 by bemath last updated on 11/Sep/20 Answered by john santu last updated on 11/Sep/20 12.3.4+13.4.5+14.5.6+15.6.7=4k=11(k+1).(k+2).(k+3)$$\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)\left({k}+\mathrm{3}\right)}=\frac{{p}}{{k}+\mathrm{1}}+\frac{{q}}{{k}+\mathrm{2}}+\frac{{r}}{{k}+\mathrm{3}} \