If-x-1-2-y-3-2-lt-1-then-find-the-range-of-x-y-x-2-y-2-

Question Number 197847 by CrispyXYZ last updated on 01/Oct/23

If (x−1)^2 +(y−(√3))^2 <1, then find the range of ((x+y)/( (√(x^2 +y^2 )))).

$$\mathrm{If}\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} <\mathrm{1},\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of} \\ $$$$\frac{{x}+{y}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}. \\ $$

Answered by Frix last updated on 01/Oct/23

In polar coordinates: ((x+y)/( (√(x^2 +y^2 ))))=cos θ +sin θ ∈[−(√2); (√2)] We need the angles of the tangents to (x−1)^2 +(y−(√3))^2 =1 through (0∣0). θ_1 =(π/6)∧θ_2 =(π/2) Now looking for min, max of cos θ +sin θ with (π/6)<θ<(π/2) which are easy to find ⇒ 1<((x+y)/( (√(x^2 +y^2 ))))≤(√2)

$$\mathrm{In}\:\mathrm{polar}\:\mathrm{coordinates}: \\ $$$$\frac{{x}+{y}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}=\mathrm{cos}\:\theta\:+\mathrm{sin}\:\theta\:\in\left[−\sqrt{\mathrm{2}};\:\sqrt{\mathrm{2}}\right] \\ $$$$\mathrm{We}\:\mathrm{need}\:\mathrm{the}\:\mathrm{angles}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tangents}\:\mathrm{to} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{1}\:\mathrm{through}\:\left(\mathrm{0}\mid\mathrm{0}\right). \\ $$$$\theta_{\mathrm{1}} =\frac{\pi}{\mathrm{6}}\wedge\theta_{\mathrm{2}} =\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{Now}\:\mathrm{looking}\:\mathrm{for}\:\mathrm{min},\:\mathrm{max}\:\mathrm{of}\:\mathrm{cos}\:\theta\:+\mathrm{sin}\:\theta \\ $$$$\mathrm{with}\:\frac{\pi}{\mathrm{6}}<\theta<\frac{\pi}{\mathrm{2}}\:\mathrm{which}\:\mathrm{are}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{find}\:\Rightarrow \\ $$$$\mathrm{1}<\frac{{x}+{y}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\leqslant\sqrt{\mathrm{2}} \\ $$

Answered by mr W last updated on 01/Oct/23

q=(1,1) p=(x,y) cos θ=((p∙q)/(∣p∣∣q∣))=((x+y)/( (√2)(√(x^2 +y^2 )))) ⇒((x+y)/( (√(x^2 +y^2 ))))=(√2) cos θ 0≤θ<(π/4) ⇒((x+y)/( (√(x^2 +y^2 )))) ∈(1,(√2)]

$$\boldsymbol{{q}}=\left(\mathrm{1},\mathrm{1}\right) \\ $$$$\boldsymbol{{p}}=\left({x},{y}\right) \\ $$$$\mathrm{cos}\:\theta=\frac{\boldsymbol{{p}}\centerdot\boldsymbol{{q}}}{\mid\boldsymbol{{p}}\mid\mid\boldsymbol{{q}}\mid}=\frac{{x}+{y}}{\:\sqrt{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }} \\ $$$$\Rightarrow\frac{{x}+{y}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}=\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{0}\leqslant\theta<\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\frac{{x}+{y}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\:\in\left(\mathrm{1},\sqrt{\mathrm{2}}\right] \\ $$

Commented by mr W last updated on 01/Oct/23