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P-Q-are-two-points-lie-on-curve-tan-2-x-y-cos-2-x-y-y-2-2y-0-on-XY-plane-If-d-PQ-then-cos-d-ans-2npi-n-N-




Question Number 63643 by vajpaithegrate@gmail.com last updated on 06/Jul/19
P(α,β) Q(γ,δ) are two points lie on curve  tan^2 (x+y)+cos^2 (x+y)+y^2 +2y=0 on  XY plane.If d=PQ then cos d=  ans:±2nπ,n∈N
$$\mathrm{P}\left(\alpha,\beta\right)\:\mathrm{Q}\left(\gamma,\delta\right)\:\mathrm{are}\:\mathrm{two}\:\mathrm{points}\:\mathrm{lie}\:\mathrm{on}\:\mathrm{curve} \\ $$$$\mathrm{tan}^{\mathrm{2}} \left(\mathrm{x}+\mathrm{y}\right)+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}+\mathrm{y}\right)+\mathrm{y}^{\mathrm{2}} +\mathrm{2y}=\mathrm{0}\:\mathrm{on} \\ $$$$\mathrm{XY}\:\mathrm{plane}.\mathrm{If}\:\mathrm{d}=\mathrm{PQ}\:\mathrm{then}\:\mathrm{cos}\:\mathrm{d}= \\ $$$$\mathrm{ans}:\pm\mathrm{2n}\pi,\mathrm{n}\in\mathrm{N} \\ $$
Answered by MJS last updated on 06/Jul/19
(d/dy)[y^2 +2y]=2y+2=0 ⇒ y=−1 ⇒  ⇒ y^2 +2y has its minimum at  (((−1)),((−1)) )  (d/dt)[tan^2  t +cos^2  t]=2((tan t)/(cos^2  t))−2sin t cos t =0 ⇒  ⇒ t=nπ  ⇒ tan^2  t +cos^2  t  has its minima at  (((nπ)),(1) )  ⇒ the only possibilities for pairs  ((x),(y) ) are  (((nπ+1)),((−1)) )  two points  (((mπ+1)),((−1)) ) and  (((nπ+1)),((−1)) ) have the  distance d=∣m−n∣π=kπ; m, n∈Z, k ∈N  ⇒ cos d =±1
$$\frac{{d}}{{dy}}\left[{y}^{\mathrm{2}} +\mathrm{2}{y}\right]=\mathrm{2}{y}+\mathrm{2}=\mathrm{0}\:\Rightarrow\:{y}=−\mathrm{1}\:\Rightarrow \\ $$$$\Rightarrow\:{y}^{\mathrm{2}} +\mathrm{2}{y}\:\mathrm{has}\:\mathrm{its}\:\mathrm{minimum}\:\mathrm{at}\:\begin{pmatrix}{−\mathrm{1}}\\{−\mathrm{1}}\end{pmatrix} \\ $$$$\frac{{d}}{{dt}}\left[\mathrm{tan}^{\mathrm{2}} \:{t}\:+\mathrm{cos}^{\mathrm{2}} \:{t}\right]=\mathrm{2}\frac{\mathrm{tan}\:{t}}{\mathrm{cos}^{\mathrm{2}} \:{t}}−\mathrm{2sin}\:{t}\:\mathrm{cos}\:{t}\:=\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:{t}={n}\pi \\ $$$$\Rightarrow\:\mathrm{tan}^{\mathrm{2}} \:{t}\:+\mathrm{cos}^{\mathrm{2}} \:{t}\:\:\mathrm{has}\:\mathrm{its}\:\mathrm{minima}\:\mathrm{at}\:\begin{pmatrix}{{n}\pi}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{only}\:\mathrm{possibilities}\:\mathrm{for}\:\mathrm{pairs}\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:\mathrm{are}\:\begin{pmatrix}{{n}\pi+\mathrm{1}}\\{−\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{two}\:\mathrm{points}\:\begin{pmatrix}{{m}\pi+\mathrm{1}}\\{−\mathrm{1}}\end{pmatrix}\:\mathrm{and}\:\begin{pmatrix}{{n}\pi+\mathrm{1}}\\{−\mathrm{1}}\end{pmatrix}\:\mathrm{have}\:\mathrm{the} \\ $$$$\mathrm{distance}\:{d}=\mid{m}−{n}\mid\pi={k}\pi;\:{m},\:{n}\in\mathbb{Z},\:{k}\:\in\mathbb{N} \\ $$$$\Rightarrow\:\mathrm{cos}\:{d}\:=\pm\mathrm{1} \\ $$

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