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Question Number 48121 by JDlix last updated on 19/Nov/18

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 ??

$${can}\:{the}\:{directrix}\:{of}\:{a}\:{parabola}\:{be}\:{in}\:{the}\:{form}\:{y}={mx}+{b}\:\:? \\ $$$${or}\:{is}\:{there}\:{an}\:{inclined}\:{parabola}\:{with}\:{directrix}\:{and}\:{axis}\: \\ $$$${of}\:{symmetry}\:{in}\:{the}\:{form}\:{of}\:{y}={mx}+{b}\:\:?? \\ $$

Commented by MJS last updated on 19/Nov/18

you can rotate a parabola, so this is indeed possible

$$\mathrm{you}\:\mathrm{can}\:\mathrm{rotate}\:\mathrm{a}\:\mathrm{parabola},\:\mathrm{so}\:\mathrm{this}\:\mathrm{is}\:\mathrm{indeed} \\ $$$$\mathrm{possible} \\ $$

Commented by JDlix last updated on 19/Nov/18

can i have example ?

$${can}\:{i}\:{have}\:{example}\:? \\ $$$$ \\ $$

Commented by MJS last updated on 20/Nov/18

y=(√3)x+((13(√3))/8)±2(√(2x+4)) is a parabola with axis y=(√3)x+((5(√3))/8) its equation is 192x^2 −128(√3)xy+64y^2 +112x−208(√3)y−517=0 the angle of the rotation: tan 2α =(B/(A−C)) ⇒ ⇒ α=−30° x=x′cos α −y′sin α =((√3)/2)x′+(1/2)y′ y=x′sin α +y′cos α =−(1/2)x′+((√3)/2)y′ ⇒ new equation is 256x^2 +160(√3)x−256y−517=0 y=x^2 +((5(√3))/8)x−((517)/(256)) y=(x+((5(√3))/(16)))^2 −((37)/(16)) ⇒ parabola y=x^2 had been shifted ((5(√3))/(16)) to the left and ((37)/(16)) down and rotated by −30°

$${y}=\sqrt{\mathrm{3}}{x}+\frac{\mathrm{13}\sqrt{\mathrm{3}}}{\mathrm{8}}\pm\mathrm{2}\sqrt{\mathrm{2}{x}+\mathrm{4}} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{parabola}\:\mathrm{with}\:\mathrm{axis}\:{y}=\sqrt{\mathrm{3}}{x}+\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$$$\mathrm{its}\:\mathrm{equation}\:\mathrm{is} \\ $$$$\mathrm{192}{x}^{\mathrm{2}} −\mathrm{128}\sqrt{\mathrm{3}}{xy}+\mathrm{64}{y}^{\mathrm{2}} +\mathrm{112}{x}−\mathrm{208}\sqrt{\mathrm{3}}{y}−\mathrm{517}=\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rotation}:\:\mathrm{tan}\:\mathrm{2}\alpha\:=\frac{{B}}{{A}−{C}}\:\Rightarrow \\ $$$$\Rightarrow\:\alpha=−\mathrm{30}° \\ $$$${x}={x}'\mathrm{cos}\:\alpha\:−{y}'\mathrm{sin}\:\alpha\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}'+\frac{\mathrm{1}}{\mathrm{2}}{y}' \\ $$$${y}={x}'\mathrm{sin}\:\alpha\:+{y}'\mathrm{cos}\:\alpha\:=−\frac{\mathrm{1}}{\mathrm{2}}{x}'+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{y}' \\ $$$$\Rightarrow\:\mathrm{new}\:\mathrm{equation}\:\mathrm{is} \\ $$$$\mathrm{256}{x}^{\mathrm{2}} +\mathrm{160}\sqrt{\mathrm{3}}{x}−\mathrm{256}{y}−\mathrm{517}=\mathrm{0} \\ $$$${y}={x}^{\mathrm{2}} +\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{8}}{x}−\frac{\mathrm{517}}{\mathrm{256}} \\ $$$${y}=\left({x}+\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{16}}\right)^{\mathrm{2}} −\frac{\mathrm{37}}{\mathrm{16}} \\ $$$$\Rightarrow\:\mathrm{parabola}\:{y}={x}^{\mathrm{2}} \:\mathrm{had}\:\mathrm{been}\:\mathrm{shifted}\:\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{16}}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{left}\:\mathrm{and}\:\frac{\mathrm{37}}{\mathrm{16}}\:\mathrm{down}\:\mathrm{and}\:\mathrm{rotated}\:\mathrm{by}\:−\mathrm{30}° \\ $$

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