Question Number 28384 by ajfour last updated on 25/Jan/18
Answered by ajfour last updated on 25/Jan/18
![If new coordinate axes be tangent at A (towards right) x axis and axis of parabola the y axis (upwards), then eq. of parabola is y=px^2 coordinates of A, O, and B are A(0,0) , O(−asin θ,−acos θ) , B[(b−a)sin θ, (b−a)cos θ] point B is on parabola, so p=((cos θ)/((b−a)sin^2 θ)) If y=mx+c is tangent to y=px^2 , we have px^2 −mx−c=0 m^2 +4cp=0 ⇒ c=−m^2 /4p eq. of a line of slope m passing through (α,β) and tangent to parabola has the form y−β=m(x−α) ⇒ β−mα=−m^2 /4p or m^2 −4mpα+4pβ=0 ∣tan φ∣=((√((m_1 +m_2 )^2 −4m_1 m_2 ))/(1+m_1 m_2 )) =((√(16α^2 p^2 −16βp))/(1+4pβ)) substituting α=−asin θ, β=−acos θ , and p=((cos θ)/((b−a)sin^2 θ)) ∣tan 𝛗∣=((4((√(((a^2 cos^2 θ)/((b−a)^2 sin^2 θ))+((acos^2 θ)/((b−a)sin^2 θ)))) ))/(∣1−((4acos^2 θ)/((b−a)sin^2 θ))∣)) 𝛗 < (π/2) for θ > tan^(−1) (((4a)/(b−a))) (if a=b, p just goes undefined ) say for θ=(π/3) ∣tan φ∣=(((4/( (√3)))(((√(ab))/(b−a))))/(∣1−((4a)/(3(b−a)))∣))=((12(√(ab)))/( (√3)∣3b−7a∣)) .](https://www.tinkutara.com/question/Q28385.png)
$${If}\:{new}\:{coordinate}\:{axes}\:{be} \\ $$$${tangent}\:{at}\:{A}\:\left({towards}\:{right}\right) \\ $$$${x}\:{axis}\:{and}\:{axis}\:{of}\:{parabola}\:{the} \\ $$$${y}\:{axis}\:\left({upwards}\right),\:{then}\:{eq}.\:{of} \\ $$$${parabola}\:{is}\:\:\boldsymbol{{y}}=\boldsymbol{{px}}^{\mathrm{2}} \\ $$$${coordinates}\:{of}\:{A},\:{O},\:{and}\:{B}\:{are} \\ $$$${A}\left(\mathrm{0},\mathrm{0}\right)\:\:,\:{O}\left(−{a}\mathrm{sin}\:\theta,−{a}\mathrm{cos}\:\theta\right)\:, \\ $$$${B}\left[\left({b}−{a}\right)\mathrm{sin}\:\theta,\:\left({b}−{a}\right)\mathrm{cos}\:\theta\right] \\ $$$${point}\:{B}\:{is}\:{on}\:{parabola},\:{so} \\ $$$$\boldsymbol{{p}}=\frac{\mathrm{cos}\:\theta}{\left({b}−{a}\right)\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$$${If}\:{y}={mx}+{c}\:\:\:\:\:\:{is}\:{tangent}\:{to} \\ $$$${y}={px}^{\mathrm{2}} \:\:,\:{we}\:{have} \\ $$$${px}^{\mathrm{2}} −{mx}−{c}=\mathrm{0} \\ $$$${m}^{\mathrm{2}} +\mathrm{4}{cp}=\mathrm{0}\:\Rightarrow\:\:\:\:{c}=−{m}^{\mathrm{2}} /\mathrm{4}{p} \\ $$$${eq}.\:{of}\:{a}\:{line}\:{of}\:{slope}\:\boldsymbol{{m}}\:{passing} \\ $$$${through}\:\left(\alpha,\beta\right)\:{and}\:{tangent}\:{to} \\ $$$${parabola}\:{has}\:{the}\:{form} \\ $$$${y}−\beta={m}\left({x}−\alpha\right) \\ $$$$\Rightarrow\:\:\:\:\beta−{m}\alpha=−{m}^{\mathrm{2}} /\mathrm{4}{p} \\ $$$${or}\:\:\:\:{m}^{\mathrm{2}} −\mathrm{4}{mp}\alpha+\mathrm{4}{p}\beta=\mathrm{0} \\ $$$$\mid\mathrm{tan}\:\phi\mid=\frac{\sqrt{\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{m}_{\mathrm{1}} {m}_{\mathrm{2}} }}{\mathrm{1}+{m}_{\mathrm{1}} {m}_{\mathrm{2}} } \\ $$$$\:\:\:\:=\frac{\sqrt{\mathrm{16}\alpha^{\mathrm{2}} {p}^{\mathrm{2}} −\mathrm{16}\beta{p}}}{\mathrm{1}+\mathrm{4}{p}\beta} \\ $$$$\:\:\:{substituting}\:\alpha=−{a}\mathrm{sin}\:\theta,\: \\ $$$$\beta=−{a}\mathrm{cos}\:\theta\:,\:{and}\:{p}=\frac{\mathrm{cos}\:\theta}{\left({b}−{a}\right)\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$$$\mid\mathrm{tan}\:\boldsymbol{\phi}\mid=\frac{\mathrm{4}\left(\sqrt{\frac{{a}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta}{\left({b}−{a}\right)^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}+\frac{{a}\mathrm{cos}\:^{\mathrm{2}} \theta}{\left({b}−{a}\right)\mathrm{sin}\:^{\mathrm{2}} \theta}}\:\right)}{\mid\mathrm{1}−\frac{\mathrm{4}{a}\mathrm{cos}\:^{\mathrm{2}} \theta}{\left({b}−{a}\right)\mathrm{sin}\:^{\mathrm{2}} \theta}\mid}\: \\ $$$$\boldsymbol{\phi}\:<\:\frac{\pi}{\mathrm{2}}\:\:{for}\:\:\theta\:>\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{4}{a}}{{b}−{a}}\right) \\ $$$$\left({if}\:{a}={b},\:\:{p}\:\:\:{just}\:{goes}\:{undefined}\:\right) \\ $$$${say}\:{for}\:\theta=\frac{\pi}{\mathrm{3}} \\ $$$$\mid\mathrm{tan}\:\phi\mid=\frac{\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\left(\frac{\sqrt{{ab}}}{{b}−{a}}\right)}{\mid\mathrm{1}−\frac{\mathrm{4}{a}}{\mathrm{3}\left({b}−{a}\right)}\mid}=\frac{\mathrm{12}\sqrt{{ab}}}{\:\sqrt{\mathrm{3}}\mid\mathrm{3}{b}−\mathrm{7}{a}\mid}\:. \\ $$
Answered by mrW2 last updated on 25/Jan/18
Commented by mrW2 last updated on 25/Jan/18
$${in}\:{coordinate}\:{system}\:{shown}\:{above}, \\ $$$${eqn}.\:{of}\:{parabola}\:{is} \\ $$$${y}={px}^{\mathrm{2}} +{q} \\ $$$${x}_{{A}} =\mathrm{0} \\ $$$${y}_{{A}} ={a}\:\mathrm{cos}\:\theta \\ $$$${x}_{{B}} =\left({b}−{a}\right)\:\mathrm{sin}\:\theta \\ $$$${y}_{{B}} ={b}\:\mathrm{cos}\:\theta \\ $$$$ \\ $$$${a}\:\mathrm{cos}\:\theta={q} \\ $$$${b}\:\mathrm{cos}\:\theta={p}\:\left({b}−{a}\right)^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta+{q} \\ $$$$\Rightarrow{q}={a}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow{p}=\frac{\mathrm{cos}\:\theta}{\left({b}−{a}\right)\:\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$ \\ $$$${x}_{{O}} =−{a}\:\mathrm{sin}\:\theta \\ $$$${y}_{{O}} =\mathrm{0} \\ $$$${eqn}.\:{of}\:{line}\:{OP}: \\ $$$${y}={m}\left({x}+{a}\:\mathrm{sin}\:\theta\right) \\ $$$${on}\:{the}\:{other}\:{side}\:{y}={px}^{\mathrm{2}} +{q} \\ $$$$\Rightarrow{px}^{\mathrm{2}} +{q}={m}\left({x}+{a}\:\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow{px}^{\mathrm{2}} −{mx}+{q}−{ma}\:\mathrm{sin}\:\theta=\mathrm{0} \\ $$$$\Delta={m}^{\mathrm{2}} −\mathrm{4}{p}\left({q}−{ma}\:\mathrm{sin}\:\theta\right)=\mathrm{0} \\ $$$$\Rightarrow{m}^{\mathrm{2}} +\mathrm{4}{pa}\:\mathrm{sin}\:\theta\:{m}−\mathrm{4}{pq}=\mathrm{0} \\ $$$$\Rightarrow{m}=\frac{−\mathrm{4}{pa}\:\mathrm{sin}\:\theta\pm\sqrt{\left(\mathrm{4}{pa}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} +\mathrm{4}×\mathrm{4}{pq}}}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{tan}\:\phi_{\mathrm{1}} ={m}_{\mathrm{1}} =−\mathrm{2}{pa}\:\mathrm{sin}\:\theta+\mathrm{2}\sqrt{\left({pa}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} +{pq}} \\ $$$$\mathrm{tan}\:\phi_{\mathrm{2}} ={m}_{\mathrm{2}} =−\mathrm{2}{pa}\:\mathrm{sin}\:\theta−\mathrm{2}\sqrt{\left({pa}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} +{pq}} \\ $$$$\mathrm{tan}\:\phi=\mathrm{tan}\:\left(\phi_{\mathrm{2}} −\phi_{\mathrm{1}} \right)=\frac{{m}_{\mathrm{2}} −{m}_{\mathrm{1}} }{\mathrm{1}+{m}_{\mathrm{2}} {m}_{\mathrm{1}} }=\frac{−\mathrm{4}\sqrt{\left({pa}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} +{pq}}}{\mathrm{1}−\mathrm{4}{pq}} \\ $$$$\mathrm{tan}\:\phi=\frac{\mathrm{4}\sqrt{\left({pa}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} +{pq}}}{\mathrm{4}{pq}−\mathrm{1}} \\ $$$$\Rightarrow\mathrm{tan}\:\phi=\frac{\mathrm{4}\sqrt{\left(\frac{{a}}{\left({b}−{a}\right)\mathrm{tan}\:\theta}\right)^{\mathrm{2}} +\frac{{a}}{\left({b}−{a}\right)\mathrm{tan}^{\mathrm{2}} \:\theta}}}{\frac{\mathrm{4}{a}}{\left({b}−{a}\right)\mathrm{tan}^{\mathrm{2}} \:\theta}−\mathrm{1}} \\ $$$$\Rightarrow\mathrm{tan}\:\phi=\frac{\mathrm{4tan}\:\theta\sqrt{{ab}}}{\mathrm{4}{a}−\left({b}−{a}\right)\mathrm{tan}^{\mathrm{2}} \:\theta} \\ $$$$\Rightarrow\phi=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4tan}\:\theta\:\sqrt{{ab}}}{\mathrm{4}{a}−\left({b}−{a}\right)\mathrm{tan}^{\mathrm{2}} \:\theta} \\ $$
Commented by ajfour last updated on 25/Jan/18
$${Thank}\:{you}\:{Sir}.\:\mathscr{E}{xcellent}! \\ $$