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Question-42863

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Question Number 42863 by ajfour last updated on 03/Sep/18
Commented by ajfour last updated on 04/Sep/18
The two parabolas have a common vertex and have the same distance from vertex to focus ′a′. Find the angle 𝛉 between their axes if their branches that are arrow marked also meet at infinity.
$${The}\:{two}\:{parabolas}\:{have}\:{a}\:{common} \\ $$$${vertex}\:{and}\:{have}\:{the}\:{same}\:{distance} \\ $$$${from}\:{vertex}\:{to}\:{focus}\:'\boldsymbol{{a}}'. \\ $$$${Find}\:{the}\:{angle}\:\boldsymbol{\theta}\:{between}\:{their} \\ $$$${axes}\:{if}\:{their}\:{branches}\:{that}\:{are} \\ $$$${arrow}\:{marked}\:{also}\:{meet}\:{at}\:{infinity}. \\ $$
Commented by MJS last updated on 04/Sep/18
they will always intersect in a real point except θ=180°
$$\mathrm{they}\:\mathrm{will}\:\mathrm{always}\:\mathrm{intersect}\:\mathrm{in}\:\mathrm{a}\:\mathrm{real}\:\mathrm{point}\:\mathrm{except} \\ $$$$\theta=\mathrm{180}° \\ $$
Commented by MrW3 last updated on 04/Sep/18
what does it mean “they meet at infinity also”?
$${what}\:{does}\:{it}\:{mean}\:“{they}\:{meet}\:{at}\:{infinity} \\ $$$${also}''? \\ $$
Commented by ajfour last updated on 04/Sep/18
they already intersect at the vertex. So that they meet also at x→∞, find θ .
$${they}\:{already}\:{intersect}\:{at}\:{the}\:{vertex}. \\ $$$${So}\:{that}\:{they}\:{meet}\:{also}\:{at}\:{x}\rightarrow\infty, \\ $$$${find}\:\theta\:. \\ $$
Commented by MJS last updated on 04/Sep/18
I guess the idea is, the left branch of the red one meets the right branch of the blue one “at infinity”, like we might think the branches of a hyperbola cross their ways “at infinity”
$$\mathrm{I}\:\mathrm{guess}\:\mathrm{the}\:\mathrm{idea}\:\mathrm{is},\:\mathrm{the}\:\mathrm{left}\:\mathrm{branch}\:\mathrm{of}\:\mathrm{the}\:\mathrm{red} \\ $$$$\mathrm{one}\:\mathrm{meets}\:\mathrm{the}\:\mathrm{right}\:\mathrm{branch}\:\mathrm{of}\:\mathrm{the}\:\mathrm{blue}\:\mathrm{one} \\ $$$$“\mathrm{at}\:\mathrm{infinity}'',\:\mathrm{like}\:\mathrm{we}\:\mathrm{might}\:\mathrm{think}\:\mathrm{the}\:\mathrm{branches} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{hyperbola}\:\mathrm{cross}\:\mathrm{their}\:\mathrm{ways}\:“\mathrm{at}\:\mathrm{infinity}'' \\ $$
Commented by MJS last updated on 04/Sep/18
think it this way: start with θ=90° and then slowly turn it counterclockwise towards θ=0° we′ll always get a real intersection except at θ=0° where the red branch “loses contact” of the right blue one but at the same moment becomes one with the left one...
$$\mathrm{think}\:\mathrm{it}\:\mathrm{this}\:\mathrm{way}: \\ $$$$\mathrm{start}\:\mathrm{with}\:\theta=\mathrm{90}°\:\mathrm{and}\:\mathrm{then}\:\mathrm{slowly}\:\mathrm{turn}\:\mathrm{it} \\ $$$$\mathrm{counterclockwise}\:\mathrm{towards}\:\theta=\mathrm{0}° \\ $$$$\mathrm{we}'\mathrm{ll}\:\mathrm{always}\:\mathrm{get}\:\mathrm{a}\:\mathrm{real}\:\mathrm{intersection}\:\mathrm{except} \\ $$$$\mathrm{at}\:\theta=\mathrm{0}°\:\mathrm{where}\:\mathrm{the}\:\mathrm{red}\:\mathrm{branch}\:“\mathrm{loses}\:\mathrm{contact}'' \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{right}\:\mathrm{blue}\:\mathrm{one}\:\mathrm{but}\:\mathrm{at}\:\mathrm{the}\:\mathrm{same}\:\mathrm{moment} \\ $$$$\mathrm{becomes}\:\mathrm{one}\:\mathrm{with}\:\mathrm{the}\:\mathrm{left}\:\mathrm{one}… \\ $$
Commented by MJS last updated on 04/Sep/18
to see what happens at infinity mirror the plain at the circle x^2 +y^2 =1 P=(r, θ) → P^∗ =((1/r), θ) ((t),((f(t))) ) → (((t/(t^2 +f^2 (t)))),(((f(t))/(t^2 +f^2 (t)))) ) a hyperbola turns into a lemniscate, it has got 2 directions (=the directions of the asymptotes) a parabola turns into a round shape with one tip in the center ((0),(0) ), it has got only one direction, its axis of symmetry. you can imagine it as an ellipse with the 2^(nd) focus in infinity
$$\mathrm{to}\:\mathrm{see}\:\mathrm{what}\:\mathrm{happens}\:\mathrm{at}\:\mathrm{infinity}\:\mathrm{mirror}\:\mathrm{the} \\ $$$$\mathrm{plain}\:\mathrm{at}\:\mathrm{the}\:\mathrm{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1} \\ $$$${P}=\left({r},\:\theta\right)\:\rightarrow\:{P}^{\ast} =\left(\frac{\mathrm{1}}{{r}},\:\theta\right) \\ $$$$\begin{pmatrix}{{t}}\\{{f}\left({t}\right)}\end{pmatrix}\:\rightarrow\:\begin{pmatrix}{\frac{{t}}{{t}^{\mathrm{2}} +{f}^{\mathrm{2}} \left({t}\right)}}\\{\frac{{f}\left({t}\right)}{{t}^{\mathrm{2}} +{f}^{\mathrm{2}} \left({t}\right)}}\end{pmatrix} \\ $$$$\mathrm{a}\:\mathrm{hyperbola}\:\mathrm{turns}\:\mathrm{into}\:\mathrm{a}\:\mathrm{lemniscate},\:\mathrm{it}\:\mathrm{has} \\ $$$$\mathrm{got}\:\mathrm{2}\:\mathrm{directions}\:\left(=\mathrm{the}\:\mathrm{directions}\:\mathrm{of}\:\mathrm{the}\right. \\ $$$$\left.\mathrm{asymptotes}\right) \\ $$$$\mathrm{a}\:\mathrm{parabola}\:\mathrm{turns}\:\mathrm{into}\:\mathrm{a}\:\mathrm{round}\:\mathrm{shape}\:\mathrm{with} \\ $$$$\mathrm{one}\:\mathrm{tip}\:\mathrm{in}\:\mathrm{the}\:\mathrm{center}\:\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix},\:\mathrm{it}\:\mathrm{has}\:\mathrm{got}\:\mathrm{only}\:\mathrm{one} \\ $$$$\mathrm{direction},\:\mathrm{its}\:\mathrm{axis}\:\mathrm{of}\:\mathrm{symmetry}.\:\mathrm{you}\:\mathrm{can}\:\mathrm{imagine} \\ $$$$\mathrm{it}\:\mathrm{as}\:\mathrm{an}\:\mathrm{ellipse}\:\mathrm{with}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{focus}\:\mathrm{in}\:\mathrm{infinity} \\ $$
Answered by MJS last updated on 04/Sep/18
y=ax^2 x → xcos θ −ysin θ y → xsin θ +ycos θ y=ax^2 → y=(x/(tan θ))+((cos θ)/(a(1−cos 2θ)))±((√(2(8axsin θ +cos 2θ +1)))/(2a(1−cos 2θ))) and this is defined for x≥−((1+cos 2θ)/(8asin θ)) with θ>0 so no matter how small ∣θ∣ is, the left branch of the red parabola will be defined in [−((1+cos 2θ)/(8asin θ)); +∞[ ⇒ it will intersect the right branch of the blue parabola
$${y}={ax}^{\mathrm{2}} \\ $$$${x}\:\rightarrow\:{x}\mathrm{cos}\:\theta\:−{y}\mathrm{sin}\:\theta \\ $$$${y}\:\rightarrow\:{x}\mathrm{sin}\:\theta\:+{y}\mathrm{cos}\:\theta \\ $$$${y}={ax}^{\mathrm{2}} \:\rightarrow\:{y}=\frac{{x}}{\mathrm{tan}\:\theta}+\frac{\mathrm{cos}\:\theta}{{a}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right)}\pm\frac{\sqrt{\mathrm{2}\left(\mathrm{8}{ax}\mathrm{sin}\:\theta\:+\mathrm{cos}\:\mathrm{2}\theta\:+\mathrm{1}\right)}}{\mathrm{2}{a}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right)} \\ $$$$\mathrm{and}\:\mathrm{this}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\geqslant−\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{8}{a}\mathrm{sin}\:\theta}\:\mathrm{with}\:\theta>\mathrm{0} \\ $$$$\mathrm{so}\:\mathrm{no}\:\mathrm{matter}\:\mathrm{how}\:\mathrm{small}\:\mid\theta\mid\:\mathrm{is},\:\mathrm{the}\:\mathrm{left}\:\mathrm{branch} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{red}\:\mathrm{parabola}\:\mathrm{will}\:\mathrm{be}\:\mathrm{defined}\:\mathrm{in} \\ $$$$\left[−\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{8}{a}\mathrm{sin}\:\theta};\:+\infty\left[\:\Rightarrow\:\mathrm{it}\:\mathrm{will}\:\mathrm{intersect}\:\mathrm{the}\:\mathrm{right}\right.\right. \\ $$$$\mathrm{branch}\:\mathrm{of}\:\mathrm{the}\:\mathrm{blue}\:\mathrm{parabola} \\ $$
Answered by MrW3 last updated on 04/Sep/18
let′s say eqn. of original parabola is y=cx^2 with c=(1/(4a)) the eqn. of rotated parabola is x sin θ+y cos θ=c (x cos θ−y sin θ)^2 intersection point (h,k): h sin θ+ch^2 cos θ=c (h cos θ−ch^2 sin θ)^2 sin θ+ch cos θ=ch (cos θ−ch sin θ)^2 with λ=ch≠0 sin θ+λ cos θ (1−cos θ)+2λ^2 sin θ cos θ−λ^3 sin^2 θ=0 sin^2 θ=((sin θ)/λ^3 )+ ((cos θ (1−cos θ))/λ^2 )+((sin 2θ)/λ) λ→∞ ⇒ sin θ→0 i.e. such that the parabolas meet at infinity, θ must equal 0.
$${let}'{s}\:{say}\:{eqn}.\:{of}\:{original}\:{parabola}\:{is} \\ $$$${y}={cx}^{\mathrm{2}} \:{with}\:{c}=\frac{\mathrm{1}}{\mathrm{4}{a}} \\ $$$${the}\:{eqn}.\:{of}\:{rotated}\:{parabola}\:{is} \\ $$$${x}\:\mathrm{sin}\:\theta+{y}\:\mathrm{cos}\:\theta={c}\:\left({x}\:\mathrm{cos}\:\theta−{y}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} \\ $$$$ \\ $$$${intersection}\:{point}\:\left({h},{k}\right): \\ $$$${h}\:\mathrm{sin}\:\theta+{ch}^{\mathrm{2}} \:\mathrm{cos}\:\theta={c}\:\left({h}\:\mathrm{cos}\:\theta−{ch}^{\mathrm{2}} \:\mathrm{sin}\:\theta\right)^{\mathrm{2}} \\ $$$$\mathrm{sin}\:\theta+{ch}\:\mathrm{cos}\:\theta={ch}\:\left(\mathrm{cos}\:\theta−{ch}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} \\ $$$${with}\:\lambda={ch}\neq\mathrm{0} \\ $$$$\mathrm{sin}\:\theta+\lambda\:\mathrm{cos}\:\theta\:\left(\mathrm{1}−\mathrm{cos}\:\theta\right)+\mathrm{2}\lambda^{\mathrm{2}} \:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta−\lambda^{\mathrm{3}} \:\mathrm{sin}^{\mathrm{2}} \:\theta=\mathrm{0} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\theta=\frac{\mathrm{sin}\:\theta}{\lambda^{\mathrm{3}} }+\:\frac{\mathrm{cos}\:\theta\:\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\lambda^{\mathrm{2}} }+\frac{\mathrm{sin}\:\mathrm{2}\theta}{\lambda} \\ $$$$\lambda\rightarrow\infty\:\Rightarrow\:\mathrm{sin}\:\theta\rightarrow\mathrm{0} \\ $$$${i}.{e}.\:{such}\:{that}\:{the}\:{parabolas}\:{meet}\:{at}\:{infinity}, \\ $$$$\theta\:{must}\:{equal}\:\mathrm{0}. \\ $$
Commented by ajfour last updated on 04/Sep/18
Thank you Sir MjS, Sir MrW, i view it better and correct now with my mind′s eye now.All answers and comments were helpful. A parabola goes parallel to its axis much away from its vertex.
$${Thank}\:{you}\:{Sir}\:{MjS},\:{Sir}\:{MrW},\: \\ $$$${i}\:{view}\:{it}\:{better}\:{and}\:{correct}\:{now} \\ $$$${with}\:{my}\:{mind}'{s}\:{eye}\:{now}.{All}\: \\ $$$${answers}\:{and}\:{comments}\:{were} \\ $$$${helpful}. \\ $$$${A}\:{parabola}\:{goes}\:{parallel}\:{to}\:{its} \\ $$$${axis}\:{much}\:{away}\:{from}\:{its}\:{vertex}. \\ $$
Commented by MJS last updated on 04/Sep/18
as always you′re welcome
$$\mathrm{as}\:\mathrm{always}\:\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$

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