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Question Number 89415 by ajfour last updated on 17/Apr/20

Commented by ajfour last updated on 17/Apr/20

Ellipse: (x^2 /4)+y^2 =1 And if ∠OP A = 90° , Find eq. of shown parabola.

$${Ellipse}:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+{y}^{\mathrm{2}} =\mathrm{1} \\ $$$${And}\:{if}\:\:\angle{OP}\:{A}\:=\:\mathrm{90}°\:,\:{Find}\:{eq}. \\ $$$${of}\:{shown}\:{parabola}. \\ $$

Commented by Tony Lin last updated on 17/Apr/20

let P(2cosθ,sinθ) (2cosθ,sinθ)∙(2cosθ−2,sinθ)=0 3cos^2 θ−4cosθ+1=0 (3cosθ−1)(cosθ−1)=0 cosθ=(1/3) or 1(false) ⇒P((2/3), ((2(√2))/3)) let parabola y=a(x−k)^2 { ((1=ak^2 )),((((2(√2))/3)=a((2/3)−k)^2 )) :}

$${let}\:{P}\left(\mathrm{2}{cos}\theta,{sin}\theta\right) \\ $$$$\left(\mathrm{2}{cos}\theta,{sin}\theta\right)\centerdot\left(\mathrm{2}{cos}\theta−\mathrm{2},{sin}\theta\right)=\mathrm{0} \\ $$$$\mathrm{3}{cos}^{\mathrm{2}} \theta−\mathrm{4}{cos}\theta+\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{3}{cos}\theta−\mathrm{1}\right)\left({cos}\theta−\mathrm{1}\right)=\mathrm{0} \\ $$$${cos}\theta=\frac{\mathrm{1}}{\mathrm{3}}\:{or}\:\mathrm{1}\left({false}\right) \\ $$$$\Rightarrow{P}\left(\frac{\mathrm{2}}{\mathrm{3}},\:\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}\right) \\ $$$${let}\:{parabola}\:{y}={a}\left({x}−{k}\right)^{\mathrm{2}} \\ $$$$\begin{cases}{\mathrm{1}={ak}^{\mathrm{2}} }\\{\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}={a}\left(\frac{\mathrm{2}}{\mathrm{3}}−{k}\right)^{\mathrm{2}} }\end{cases} \\ $$

Commented by ajfour last updated on 17/Apr/20

this cannot be eq. of the required parabola, please solve again, Sir.

$${this}\:{cannot}\:{be}\:{eq}.\:{of}\:{the}\:{required} \\ $$$${parabola},\:{please}\:{solve}\:{again},\:{Sir}. \\ $$

Commented by Tony Lin last updated on 17/Apr/20

⇒ { ((a=(9/4)+(3/(√2))−((3(√3))/(2)^(1/4) ))),((k=(2+((4(√2))/3))(3+(√(6(√2)))))) :} ⇒y=((9/4)+(3/(√2))−((3(√3))/(2)^(1/4) )){x−(2+((4(√2))/3))(3+(√(6(√2))))}^2

$$\Rightarrow\begin{cases}{{a}=\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{3}}{\sqrt{\mathrm{2}}}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\sqrt[{\mathrm{4}}]{\mathrm{2}}}}\\{{k}=\left(\mathrm{2}+\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{3}}\right)\left(\mathrm{3}+\sqrt{\mathrm{6}\sqrt{\mathrm{2}}}\right)}\end{cases} \\ $$$$\Rightarrow{y}=\left(\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{3}}{\sqrt{\mathrm{2}}}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\sqrt[{\mathrm{4}}]{\mathrm{2}}}\right)\left\{{x}−\left(\mathrm{2}+\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{3}}\right)\left(\mathrm{3}+\sqrt{\mathrm{6}\sqrt{\mathrm{2}}}\right)\right\}^{\mathrm{2}} \\ $$

Answered by mr W last updated on 18/Apr/20

(x^2 /4)+y^2 =1 P(2 cos θ, sin θ) (2 cos θ)^2 +sin^2 θ+4(1−cos θ)^2 +sin^2 θ=4 3 cos^2 θ−4 cos θ+1=0 (3 cos θ−1)(cos θ−1)=0 ⇒cos θ=(1/3) ⇒sin θ=((2(√2))/3) P((2/3), ((2(√2))/3)) eqn. of parabola: say y=A(x−B)^2 1=AB^2 ...(i) ((2(√2))/3)=A((2/3)−B)^2 ...(ii) ((2/(3B))−1)^2 =((2(√2))/3) B=(2/(3±(√(6(√2))))) ⇒B=(2/(3+(√(6(√2)))))<(2/3) A=(1/B^2 ) eqn. : y=(1/B^2 )(x−B)^2 =((x/B)−1)^2 ⇒ y=[(((3+(√(6(√2))))x)/2)−1]^2

$$\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+{y}^{\mathrm{2}} =\mathrm{1} \\ $$$${P}\left(\mathrm{2}\:\mathrm{cos}\:\theta,\:\mathrm{sin}\:\theta\right) \\ $$$$\left(\mathrm{2}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{4}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \:\theta=\mathrm{4} \\ $$$$\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{4}\:\mathrm{cos}\:\theta+\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{3}\:\mathrm{cos}\:\theta−\mathrm{1}\right)\left(\mathrm{cos}\:\theta−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\mathrm{sin}\:\theta=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$${P}\left(\frac{\mathrm{2}}{\mathrm{3}},\:\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}\right) \\ $$$${eqn}.\:{of}\:{parabola}:\:{say}\:{y}={A}\left({x}−{B}\right)^{\mathrm{2}} \\ $$$$\mathrm{1}={AB}^{\mathrm{2}} \:\:\:...\left({i}\right) \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}={A}\left(\frac{\mathrm{2}}{\mathrm{3}}−{B}\right)^{\mathrm{2}} \:\:\:...\left({ii}\right) \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}{B}}−\mathrm{1}\right)^{\mathrm{2}} =\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$${B}=\frac{\mathrm{2}}{\mathrm{3}\pm\sqrt{\mathrm{6}\sqrt{\mathrm{2}}}} \\ $$$$\Rightarrow{B}=\frac{\mathrm{2}}{\mathrm{3}+\sqrt{\mathrm{6}\sqrt{\mathrm{2}}}}<\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${A}=\frac{\mathrm{1}}{{B}^{\mathrm{2}} } \\ $$$$ \\ $$$${eqn}.\:: \\ $$$${y}=\frac{\mathrm{1}}{{B}^{\mathrm{2}} }\left({x}−{B}\right)^{\mathrm{2}} =\left(\frac{{x}}{{B}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{y}=\left[\frac{\left(\mathrm{3}+\sqrt{\mathrm{6}\sqrt{\mathrm{2}}}\right){x}}{\mathrm{2}}−\mathrm{1}\right]^{\mathrm{2}} \\ $$

Commented by mr W last updated on 18/Apr/20

Commented by ajfour last updated on 19/Apr/20

Thanks, Sir.

$${Thanks},\:{Sir}. \\ $$

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