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Question-81854

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Question Number 81854 by ajfour last updated on 16/Feb/20
Commented by ajfour last updated on 16/Feb/20
Find eq. of parabola using a,b,c in the form y=q−x^2 .
$${Find}\:{eq}.\:{of}\:{parabola}\:{using} \\ $$$${a},{b},{c}\:\:{in}\:{the}\:{form}\:{y}={q}−{x}^{\mathrm{2}} . \\ $$
Answered by mr W last updated on 16/Feb/20
Commented by mr W last updated on 16/Feb/20
an attempt... ϕ=cos^(−1) ((a^2 +b^2 −c^2 )/(2ab)) x_B =a cos θ y_B =a sin θ ⇒a sin θ=q−a^2 cos^2 θ ...(i) x_A =b cos (θ+ϕ)=b(cos θcos ϕ−sin θsin ϕ) y_A =b sin (θ+ϕ)=b(sin θcos ϕ+cos θsin ϕ) ⇒b(sin θcos ϕ+cos θsin ϕ)=q−b^2 (cos θcos ϕ−sin θsin ϕ)^2 (ii)−(i): b(sin θcos ϕ+cos θsin ϕ)−a sin θ =a^2 cos^2 θ−b^2 (cos θcos ϕ−sin θsin ϕ)^2 we can solve this for θ and then get q from (i): q=a(sin θ+a^2 cos θ)
$${an}\:{attempt}… \\ $$$$\varphi=\mathrm{cos}^{−\mathrm{1}} \frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$${x}_{{B}} ={a}\:\mathrm{cos}\:\theta \\ $$$${y}_{{B}} ={a}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow{a}\:\mathrm{sin}\:\theta={q}−{a}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta\:\:\:…\left({i}\right) \\ $$$${x}_{{A}} ={b}\:\mathrm{cos}\:\left(\theta+\varphi\right)={b}\left(\mathrm{cos}\:\theta\mathrm{cos}\:\varphi−\mathrm{sin}\:\theta\mathrm{sin}\:\varphi\right) \\ $$$${y}_{{A}} ={b}\:\mathrm{sin}\:\left(\theta+\varphi\right)={b}\left(\mathrm{sin}\:\theta\mathrm{cos}\:\varphi+\mathrm{cos}\:\theta\mathrm{sin}\:\varphi\right) \\ $$$$\Rightarrow{b}\left(\mathrm{sin}\:\theta\mathrm{cos}\:\varphi+\mathrm{cos}\:\theta\mathrm{sin}\:\varphi\right)={q}−{b}^{\mathrm{2}} \left(\mathrm{cos}\:\theta\mathrm{cos}\:\varphi−\mathrm{sin}\:\theta\mathrm{sin}\:\varphi\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$${b}\left(\mathrm{sin}\:\theta\mathrm{cos}\:\varphi+\mathrm{cos}\:\theta\mathrm{sin}\:\varphi\right)−{a}\:\mathrm{sin}\:\theta \\ $$$$={a}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta−{b}^{\mathrm{2}} \left(\mathrm{cos}\:\theta\mathrm{cos}\:\varphi−\mathrm{sin}\:\theta\mathrm{sin}\:\varphi\right)^{\mathrm{2}} \\ $$$$ \\ $$$${we}\:{can}\:{solve}\:{this}\:{for}\:\theta\:{and}\:{then} \\ $$$$\:{get}\:{q}\:{from}\:\left({i}\right): \\ $$$${q}={a}\left(\mathrm{sin}\:\theta+{a}^{\mathrm{2}} \:\mathrm{cos}\:\theta\right) \\ $$
Commented by ajfour last updated on 16/Feb/20
equivalently let further A(bcos ψ,bsin ψ) ⇒ a^2 sin^2 θ−asin θ+(q−a^2 )=0 b^2 sin^2 ψ−bsin ψ+(q−b^2 )=0 sin θ=(1/(2a))+(√((1/(4a^2 ))−(q/a^2 )+1)) sin ψ=(1/(2b))+(√((1/(4b^2 ))−(q/b^2 )+1)) ________________________ And θ+ψ = sin^(−1) {(1/(2a))+(√((1/(4a^2 ))−(q/a^2 )+1))} +sin^(−1) {(1/(2b))+(√((1/(4b^2 ))−(q/b^2 )+1))} =π−cos^(−1) (((a^2 +b^2 −c^2 )/(2ab))) _________________________■
$${equivalently} \\ $$$${let}\:{further}\:{A}\left({b}\mathrm{cos}\:\psi,{b}\mathrm{sin}\:\psi\right) \\ $$$$\Rightarrow\:\:{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta−{a}\mathrm{sin}\:\theta+\left({q}−{a}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:{b}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \psi−{b}\mathrm{sin}\:\psi+\left({q}−{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\:\:\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{2}{a}}+\sqrt{\frac{\mathrm{1}}{\mathrm{4}{a}^{\mathrm{2}} }−\frac{{q}}{{a}^{\mathrm{2}} }+\mathrm{1}} \\ $$$$\:\:\mathrm{sin}\:\psi=\frac{\mathrm{1}}{\mathrm{2}{b}}+\sqrt{\frac{\mathrm{1}}{\mathrm{4}{b}^{\mathrm{2}} }−\frac{{q}}{{b}^{\mathrm{2}} }+\mathrm{1}} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${And}\:\:\:\theta+\psi\:=\: \\ $$$$\:\:\:\:\:\:\:\mathrm{sin}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{\mathrm{2}{a}}+\sqrt{\frac{\mathrm{1}}{\mathrm{4}{a}^{\mathrm{2}} }−\frac{{q}}{{a}^{\mathrm{2}} }+\mathrm{1}}\right\} \\ $$$$\:\:\:\:+\mathrm{sin}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{\mathrm{2}{b}}+\sqrt{\frac{\mathrm{1}}{\mathrm{4}{b}^{\mathrm{2}} }−\frac{{q}}{{b}^{\mathrm{2}} }+\mathrm{1}}\right\} \\ $$$$\:\:\:\:=\pi−\mathrm{cos}^{−\mathrm{1}} \left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\__{\blacksquare} \\ $$
Commented by mr W last updated on 16/Feb/20
very nice with a final eqn. for q!
$${very}\:{nice}\:{with}\:{a}\:{final}\:{eqn}.\:{for}\:{q}! \\ $$

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