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Geometry-What-are-the-equations-for-two-lines-through-the-origin-that-are-tangent-to-the-ellipse-6-48-5-0-

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Question Number 134951 by bobhans last updated on 09/Mar/21
Geometry What are the equations for two lines through the origin that are tangent to the ellipseΒ  6π‘₯Β²βˆ’48π‘₯+𝑦²+5=0 ?
$$\underline{\mathrm{Geometry}} \\ $$What are the equations for two lines through the origin that are tangent to the ellipseΒ  6π‘₯Β²βˆ’48π‘₯+𝑦²+5=0 ?
Answered by EDWIN88 last updated on 09/Mar/21
let the equation tangent line to ellipse 6x^2 βˆ’48x+y^2 +5 = 0 is y = mx substitute to ellipse give 6x^2 βˆ’48x+m^2 x^2 +5 = 0 (m^2 +6)x^2 βˆ’48x+5 = 0 , take Ξ” = 0 β‡’ Ξ”=48^2 βˆ’4(m^2 +6)(5) = 0 β‡’12Γ—48 = 5(m^2 +6) β‡’m^2 = ((12Γ—48)/5)βˆ’6 = ((546)/5) β‡’ m = Β± (√((546)/5)) . so equation of two line through the origin that are tangent to the ellipse 6x^2 βˆ’48x+y^2 +5 = 0 is y=Β±x(√((546)/5))
$$\mathrm{let}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{to}\:\mathrm{ellipse} \\ $$$$\mathrm{6x}^{\mathrm{2}} βˆ’\mathrm{48x}+\mathrm{y}^{\mathrm{2}} +\mathrm{5}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{y}\:=\:\mathrm{mx} \\ $$$$\mathrm{substitute}\:\mathrm{to}\:\mathrm{ellipse}\:\mathrm{give}\: \\ $$$$\mathrm{6x}^{\mathrm{2}} βˆ’\mathrm{48x}+\mathrm{m}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} +\mathrm{5}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{m}^{\mathrm{2}} +\mathrm{6}\right)\mathrm{x}^{\mathrm{2}} βˆ’\mathrm{48x}+\mathrm{5}\:=\:\mathrm{0}\:,\:\mathrm{take}\:\Delta\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\Delta=\mathrm{48}^{\mathrm{2}} βˆ’\mathrm{4}\left(\mathrm{m}^{\mathrm{2}} +\mathrm{6}\right)\left(\mathrm{5}\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{12}Γ—\mathrm{48}\:=\:\mathrm{5}\left(\mathrm{m}^{\mathrm{2}} +\mathrm{6}\right) \\ $$$$\Rightarrow\mathrm{m}^{\mathrm{2}} \:=\:\frac{\mathrm{12}Γ—\mathrm{48}}{\mathrm{5}}βˆ’\mathrm{6}\:=\:\frac{\mathrm{546}}{\mathrm{5}} \\ $$$$\Rightarrow\:\mathrm{m}\:=\:\pm\:\sqrt{\frac{\mathrm{546}}{\mathrm{5}}}\:.\:\mathrm{so}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{two}\:\mathrm{line} \\ $$$$\mathrm{through}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{that}\:\mathrm{are}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{ellipse}\:\mathrm{6x}^{\mathrm{2}} βˆ’\mathrm{48x}+\mathrm{y}^{\mathrm{2}} +\mathrm{5}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{y}=\pm\mathrm{x}\sqrt{\frac{\mathrm{546}}{\mathrm{5}}} \\ $$$$ \\ $$
Commented by EDWIN88 last updated on 09/Mar/21

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