Menu Close

Question-205594

Tinku Tara 0 Comments
Pin




Question Number 205594 by cortano12 last updated on 25/Mar/24
Answered by Red1ight last updated on 25/Mar/24
y^2 =4x y=2(√x) (dy/dx)=(1/( (√x)))=(2/y) (x_0 ,y_0 )=(2,8),m=−(√x) y=8+2(√x)−x(√x) 8+2(√x)−x(√x)=2(√x) 8−x(√x)=0 x^(3/2) =8 x=4 C=(4,4) B=(x_b ,y_b ),m=−2 x_b =2+cos(tan^(−1) (2))=2+((√5)/5) y_b =8+sin(tan^(−1) (2))=8+((−2(√5))/5) B≈(2.447,7.105) BC≈7.067
$${y}^{\mathrm{2}} =\mathrm{4}{x} \\ $$$${y}=\mathrm{2}\sqrt{{x}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\:\sqrt{{x}}}=\frac{\mathrm{2}}{{y}} \\ $$$$\left({x}_{\mathrm{0}} ,{y}_{\mathrm{0}} \right)=\left(\mathrm{2},\mathrm{8}\right),{m}=−\sqrt{{x}} \\ $$$${y}=\mathrm{8}+\mathrm{2}\sqrt{{x}}−{x}\sqrt{{x}} \\ $$$$\mathrm{8}+\mathrm{2}\sqrt{{x}}−{x}\sqrt{{x}}=\mathrm{2}\sqrt{{x}} \\ $$$$\mathrm{8}−{x}\sqrt{{x}}=\mathrm{0} \\ $$$${x}^{\frac{\mathrm{3}}{\mathrm{2}}} =\mathrm{8} \\ $$$${x}=\mathrm{4} \\ $$$${C}=\left(\mathrm{4},\mathrm{4}\right) \\ $$$${B}=\left({x}_{{b}} ,{y}_{{b}} \right),{m}=−\mathrm{2} \\ $$$${x}_{{b}} =\mathrm{2}+\mathrm{cos}\left(\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)\right)=\mathrm{2}+\frac{\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$$${y}_{{b}} =\mathrm{8}+\mathrm{sin}\left(\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)\right)=\mathrm{8}+\frac{−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$$${B}\approx\left(\mathrm{2}.\mathrm{447},\mathrm{7}.\mathrm{105}\right) \\ $$$${BC}\approx\mathrm{7}.\mathrm{067} \\ $$$$ \\ $$
Commented by Red1ight last updated on 25/Mar/24
Commented by Red1ight last updated on 25/Mar/24
Switched B and C
$$\mathrm{Switched}\:{B}\:\mathrm{and}\:{C} \\ $$
Answered by mr W last updated on 25/Mar/24
Commented by mr W last updated on 25/Mar/24
A(2,8) P(p,q) 4p=q^2 Φ=R^2 =(p−2)^2 +(q−8)^2 =((q^2 /4)−2)^2 +(q−8)^2 (dΦ/dq)=q((q^2 /4)−2)+2(q−8)=0 ⇒q^3 −64=0 ⇒q=4 ⇒p=4 ⇒R_(min) ^2 =(4−2)^2 +(4−8)^2 =20 ⇒R_(min) =2(√5) min BC=R_(min) −1=2(√5)−1≈3.472 ✓
$${A}\left(\mathrm{2},\mathrm{8}\right) \\ $$$${P}\left({p},{q}\right) \\ $$$$\mathrm{4}{p}={q}^{\mathrm{2}} \\ $$$$\Phi={R}^{\mathrm{2}} =\left({p}−\mathrm{2}\right)^{\mathrm{2}} +\left({q}−\mathrm{8}\right)^{\mathrm{2}} =\left(\frac{{q}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{2}\right)^{\mathrm{2}} +\left({q}−\mathrm{8}\right)^{\mathrm{2}} \\ $$$$\frac{{d}\Phi}{{dq}}={q}\left(\frac{{q}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{2}\right)+\mathrm{2}\left({q}−\mathrm{8}\right)=\mathrm{0} \\ $$$$\Rightarrow{q}^{\mathrm{3}} −\mathrm{64}=\mathrm{0} \\ $$$$\Rightarrow{q}=\mathrm{4}\:\Rightarrow{p}=\mathrm{4} \\ $$$$\Rightarrow{R}_{{min}} ^{\mathrm{2}} =\left(\mathrm{4}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{4}−\mathrm{8}\right)^{\mathrm{2}} =\mathrm{20}\:\Rightarrow{R}_{{min}} =\mathrm{2}\sqrt{\mathrm{5}} \\ $$$${min}\:{BC}={R}_{{min}} −\mathrm{1}=\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{1}\approx\mathrm{3}.\mathrm{472}\:\checkmark \\ $$
Commented by mr W last updated on 25/Mar/24
Answered by mr W last updated on 25/Mar/24
Method II B(2+cos θ, 8+sin θ) C((y^2 /4), y) Φ=BC^2 =(2+cos θ−(y^2 /4))^2 +(8+sin θ−y)^2 (∂Φ/∂θ)=−2sin θ(2+cos θ−(y^2 /4))+2cos θ(8+sin θ−y)=0 sin θ(2+cos θ−(y^2 /4))=cos θ(8+sin θ−y)=0 ⇒tan θ=((4(8−y))/(8−y^2 )) (∂Φ/∂y)=−y(2+cos θ−(y^2 /4))−2(8+sin θ−y)=0 −y(2+cos θ−(y^2 /4))=2(8+sin θ−y) ⇒−(y/2)=((sin θ)/(cos θ))=tan θ ((4(8−y))/(8−y^2 ))=−(y/2) y^3 =64 ⇒y=4 ⇒tan θ=−2 ⇒ { ((cos θ=−(1/( (√5))) or (1/( (√5))))),((sin θ=(2/( (√5))) or −(2/( (√5))))) :} BC_(min) ^2 =(2+(1/( (√5)))−(4^2 /4))^2 +(8−(2/( (√5)))−4)^2 =21−4(√5) ⇒(BC)_(min) =(√(21−4(√5)))=2(√5)−1 ✓
$${Method}\:{II} \\ $$$${B}\left(\mathrm{2}+\mathrm{cos}\:\theta,\:\mathrm{8}+\mathrm{sin}\:\theta\right) \\ $$$${C}\left(\frac{{y}^{\mathrm{2}} }{\mathrm{4}},\:{y}\right) \\ $$$$\Phi={BC}^{\mathrm{2}} =\left(\mathrm{2}+\mathrm{cos}\:\theta−\frac{{y}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{2}} +\left(\mathrm{8}+\mathrm{sin}\:\theta−{y}\right)^{\mathrm{2}} \\ $$$$\frac{\partial\Phi}{\partial\theta}=−\mathrm{2sin}\:\theta\left(\mathrm{2}+\mathrm{cos}\:\theta−\frac{{y}^{\mathrm{2}} }{\mathrm{4}}\right)+\mathrm{2cos}\:\theta\left(\mathrm{8}+\mathrm{sin}\:\theta−{y}\right)=\mathrm{0} \\ $$$$\mathrm{sin}\:\theta\left(\mathrm{2}+\mathrm{cos}\:\theta−\frac{{y}^{\mathrm{2}} }{\mathrm{4}}\right)=\mathrm{cos}\:\theta\left(\mathrm{8}+\mathrm{sin}\:\theta−{y}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\mathrm{4}\left(\mathrm{8}−{y}\right)}{\mathrm{8}−{y}^{\mathrm{2}} } \\ $$$$\frac{\partial\Phi}{\partial{y}}=−{y}\left(\mathrm{2}+\mathrm{cos}\:\theta−\frac{{y}^{\mathrm{2}} }{\mathrm{4}}\right)−\mathrm{2}\left(\mathrm{8}+\mathrm{sin}\:\theta−{y}\right)=\mathrm{0} \\ $$$$−{y}\left(\mathrm{2}+\mathrm{cos}\:\theta−\frac{{y}^{\mathrm{2}} }{\mathrm{4}}\right)=\mathrm{2}\left(\mathrm{8}+\mathrm{sin}\:\theta−{y}\right) \\ $$$$\Rightarrow−\frac{{y}}{\mathrm{2}}=\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}=\mathrm{tan}\:\theta \\ $$$$\frac{\mathrm{4}\left(\mathrm{8}−{y}\right)}{\mathrm{8}−{y}^{\mathrm{2}} }=−\frac{{y}}{\mathrm{2}} \\ $$$${y}^{\mathrm{3}} =\mathrm{64}\:\Rightarrow{y}=\mathrm{4} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=−\mathrm{2}\:\Rightarrow\begin{cases}{\mathrm{cos}\:\theta=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:{or}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}}\\{\mathrm{sin}\:\theta=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:{or}\:−\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}}\end{cases} \\ $$$${BC}_{{min}} ^{\mathrm{2}} =\left(\mathrm{2}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}−\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{2}} +\left(\mathrm{8}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{21}−\mathrm{4}\sqrt{\mathrm{5}} \\ $$$$\Rightarrow\left({BC}\right)_{{min}} =\sqrt{\mathrm{21}−\mathrm{4}\sqrt{\mathrm{5}}}=\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{1}\:\checkmark \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *