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Question-39470

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Question Number 39470 by Raj Singh last updated on 06/Jul/18
Commented by maxmathsup by imad last updated on 07/Jul/18
x^2 +y^2 =1 ⇒y^2 =1−x^2 ⇒1−x^2 =4x−4 ⇒x^2 +4x −5 =0⇒ Δ^′ =2^2 +5 =9 ⇒ x_1 =−2 +3 =1 and x_2 =−2−3 =−5 x=1⇒y=0 ⇒ A(1,0) is intersection x=−5 ⇒y^2 =−24 and this is impossibleso the first point is A.
$${x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:=\mathrm{1}\:\Rightarrow{y}^{\mathrm{2}} =\mathrm{1}−{x}^{\mathrm{2}} \:\Rightarrow\mathrm{1}−{x}^{\mathrm{2}} \:=\mathrm{4}{x}−\mathrm{4}\:\Rightarrow{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:−\mathrm{5}\:=\mathrm{0}\Rightarrow \\ $$$$\Delta^{'} =\mathrm{2}^{\mathrm{2}} \:\:+\mathrm{5}\:=\mathrm{9}\:\:\Rightarrow\:{x}_{\mathrm{1}} =−\mathrm{2}\:+\mathrm{3}\:=\mathrm{1}\:\:{and}\:{x}_{\mathrm{2}} =−\mathrm{2}−\mathrm{3}\:=−\mathrm{5} \\ $$$${x}=\mathrm{1}\Rightarrow{y}=\mathrm{0}\:\Rightarrow\:{A}\left(\mathrm{1},\mathrm{0}\right)\:{is}\:{intersection} \\ $$$${x}=−\mathrm{5}\:\Rightarrow{y}^{\mathrm{2}} \:=−\mathrm{24}\:\:{and}\:{this}\:{is}\:{impossibleso}\:{the}\:{first}\:{point}\:{is}\:{A}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18
Answered by ajfour last updated on 06/Jul/18
x^2 +4(x−1)=1 x^2 +4x−5=0 (x+2)^2 =9 x=−2±3 ; but for parabola x ≥ 1 hence only point is (1,0) ; hence they touch (circle being a closed curve).
$${x}^{\mathrm{2}} +\mathrm{4}\left({x}−\mathrm{1}\right)=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{5}=\mathrm{0} \\ $$$$\left({x}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$${x}=−\mathrm{2}\pm\mathrm{3}\:;\:\:{but}\:{for}\:{parabola}\:{x}\:\geqslant\:\mathrm{1} \\ $$$${hence}\:{only}\:{point}\:{is}\:\left(\mathrm{1},\mathrm{0}\right)\:;\:{hence} \\ $$$${they}\:{touch}\:\left({circle}\:{being}\:{a}\:{closed}\right. \\ $$$$\left.{curve}\right). \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18
solve... x^2 +4x−4=1 x^2 +5x−x−5=0 x(x+5)−1(x+5)=0 (x+5)(x−1)=0 x=1,−5 when x=1 y=0 25+y^2 =1 then y^2 =−24 which is not possible. a circle and parabola intesect at two distinct points But here only one solution(1,0) satisfy both equation... so thiz two curve circle and parabola can not intersect at single point...Hence they touch each other at (1,0)
$${solve}… \\ $$$${x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{4}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +\mathrm{5}{x}−{x}−\mathrm{5}=\mathrm{0} \\ $$$${x}\left({x}+\mathrm{5}\right)−\mathrm{1}\left({x}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\left({x}+\mathrm{5}\right)\left({x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{1},−\mathrm{5} \\ $$$${when}\:{x}=\mathrm{1}\:\:{y}=\mathrm{0} \\ $$$$\mathrm{25}+{y}^{\mathrm{2}} =\mathrm{1}\:\:{then}\:{y}^{\mathrm{2}} =−\mathrm{24}\:\:{which}\:{is}\:{not}\:{possible}. \\ $$$${a}\:{circle}\:{and}\:{parabola}\:{intesect}\:{at}\:{two}\:{distinct} \\ $$$${points}\:\boldsymbol{{B}}{ut}\:{here}\:{only}\:{one}\:{solution}\left(\mathrm{1},\mathrm{0}\right) \\ $$$${satisfy}\:{both}\:{equation}… \\ $$$${so}\:{thiz}\:{two}\:{curve}\:{circle}\:{and}\:{parabola}\:{can}\:{not} \\ $$$${intersect}\:{at}\:{single}\:{point}…\boldsymbol{{H}}{ence}\:{they}\:{touch} \\ $$$${each}\:{other}\:{at}\:\left(\mathrm{1},\mathrm{0}\right) \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Jul/18
when touch each other slope value caculated st that with respect to both curve should be same x^2 +y^2 =1 2x +2y.(dy/dx)=0 (dy/dx) at(1,0) is m_1 =(((−x)/y))_(1′0) =∞ y^2 =4(x−1) 2y(dy/dx)=4 (dy/dx)=(2/y) so slope at (1,0) is=∞=
$${when}\:{touch}\:\:{each}\:{other}\:{slope}\:{value}\:{caculated}\:{st}\:{that}\:{with}\: \\ $$$${respect}\:{to}\:{both}\:{curve}\:{should}\:{be}\:{same} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2}{x}\:+\mathrm{2}{y}.\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}\:{at}\left(\mathrm{1},\mathrm{0}\right)\:{is}\:\:{m}_{\mathrm{1}} =\left(\frac{−{x}}{{y}}\right)_{\mathrm{1}'\mathrm{0}} =\infty \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${y}^{\mathrm{2}} =\mathrm{4}\left({x}−\mathrm{1}\right) \\ $$$$\mathrm{2}{y}\frac{{dy}}{{dx}}=\mathrm{4}\:\:\:\:\frac{{dy}}{{dx}}=\frac{\mathrm{2}}{{y}}\:\:{so}\:{slope}\:{at}\:\left(\mathrm{1},\mathrm{0}\right)\:\:{is}=\infty= \\ $$

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