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Question Number 150429 by ZiYangLee last updated on 12/Aug/21

Find the equations of the common tangents to the parabola y^2 =4x and the parabola x^2 =2y−3.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{common} \\ $$$$\mathrm{tangents}\:\mathrm{to}\:\mathrm{the}\:\mathrm{parabola}\:{y}^{\mathrm{2}} =\mathrm{4}{x}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{parabola}\:{x}^{\mathrm{2}} =\mathrm{2}{y}−\mathrm{3}. \\ $$

Answered by Olaf_Thorendsen last updated on 12/Aug/21

If Δ is a tangent of the parabola y = (1/2)x^2 +(3/2) at x_0 : Δ : y−y_0 = (dy/dx)(x_0 ).(x−x_0 ) Δ : y−(1/2)x_0 ^2 −(3/2) = x_0 (x−x_0 ) Δ : y = x_0 x−(1/2)x_0 ^2 +(3/2) If Δ′ is a tangent of the parabola x = (1/4)y^2 at y_1 : Δ′ : x−x_1 = (dx/dy)(y_1 ).(y−y_1 ) Δ′ : x−(1/4)y_1 ^2 = (1/2)y_1 (y−y_1 ) Δ′ : x = (1/2)y_1 y−(1/4)y_1 ^2 Δ′ : y = (2/y_1 )x+(1/2)y_1 Δ and Δ′ are the same tangent if the equation is the same : x_0 = (2/y_1 ) and −(1/2)x_0 ^2 +(3/2) = (1/2)y_1 −(1/2).(4/y_1 ^2 )+(3/2) = (1/2)y_1 y_1 ^3 −3y_1 ^2 +4 = 0 (y_1 +1)(y_1 ^2 −4y_1 +4) = 0 (y_1 +1)(y_1 −2)^2 = 0 y_1 = −1 or +2 Δ′ : y = (2/y_1 )x +(1/2)y_1 Δ′ : y = −2x −(1/2) or y = x+1

$$\mathrm{If}\:\Delta\:\mathrm{is}\:\mathrm{a}\:\mathrm{tangent}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parabola} \\ $$$${y}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{at}\:{x}_{\mathrm{0}} \:: \\ $$$$\Delta\::\:{y}−{y}_{\mathrm{0}} \:=\:\frac{{dy}}{{dx}}\left({x}_{\mathrm{0}} \right).\left({x}−{x}_{\mathrm{0}} \right) \\ $$$$\Delta\::\:{y}−\frac{\mathrm{1}}{\mathrm{2}}{x}_{\mathrm{0}} ^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}\:=\:{x}_{\mathrm{0}} \left({x}−{x}_{\mathrm{0}} \right) \\ $$$$\Delta\::\:{y}\:=\:{x}_{\mathrm{0}} {x}−\frac{\mathrm{1}}{\mathrm{2}}{x}_{\mathrm{0}} ^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{If}\:\Delta'\:\mathrm{is}\:\mathrm{a}\:\mathrm{tangent}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parabola} \\ $$$${x}\:=\:\frac{\mathrm{1}}{\mathrm{4}}{y}^{\mathrm{2}} \:\mathrm{at}\:{y}_{\mathrm{1}} \:: \\ $$$$\Delta'\::\:{x}−{x}_{\mathrm{1}} \:=\:\frac{{dx}}{{dy}}\left({y}_{\mathrm{1}} \right).\left({y}−{y}_{\mathrm{1}} \right) \\ $$$$\Delta'\::\:{x}−\frac{\mathrm{1}}{\mathrm{4}}{y}_{\mathrm{1}} ^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{y}_{\mathrm{1}} \left({y}−{y}_{\mathrm{1}} \right) \\ $$$$\Delta'\::\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{y}_{\mathrm{1}} {y}−\frac{\mathrm{1}}{\mathrm{4}}{y}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\Delta'\::\:{y}\:=\:\frac{\mathrm{2}}{{y}_{\mathrm{1}} }{x}+\frac{\mathrm{1}}{\mathrm{2}}{y}_{\mathrm{1}} \\ $$$$ \\ $$$$\Delta\:\mathrm{and}\:\Delta'\:\mathrm{are}\:\mathrm{the}\:\mathrm{same}\:\mathrm{tangent}\:\mathrm{if}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:: \\ $$$${x}_{\mathrm{0}} \:=\:\frac{\mathrm{2}}{{y}_{\mathrm{1}} }\:\mathrm{and}\:−\frac{\mathrm{1}}{\mathrm{2}}{x}_{\mathrm{0}} ^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{y}_{\mathrm{1}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{4}}{{y}_{\mathrm{1}} ^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{y}_{\mathrm{1}} \\ $$$${y}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{3}{y}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{4}\:=\:\mathrm{0} \\ $$$$\left({y}_{\mathrm{1}} +\mathrm{1}\right)\left({y}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{4}{y}_{\mathrm{1}} +\mathrm{4}\right)\:=\:\mathrm{0} \\ $$$$\left({y}_{\mathrm{1}} +\mathrm{1}\right)\left({y}_{\mathrm{1}} −\mathrm{2}\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$${y}_{\mathrm{1}} \:=\:−\mathrm{1}\:\mathrm{or}\:+\mathrm{2} \\ $$$$\Delta'\::\:{y}\:=\:\frac{\mathrm{2}}{{y}_{\mathrm{1}} }{x}\:+\frac{\mathrm{1}}{\mathrm{2}}{y}_{\mathrm{1}} \\ $$$$\Delta'\::\:{y}\:=\:−\mathrm{2}{x}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{or}\:{y}\:=\:{x}+\mathrm{1} \\ $$

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