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Question Number 106408 by bemath last updated on 05/Aug/20
what is k such that the circle x^2 +y^2 =4 intersect the parabola y= x^2 +k
whatisksuchthatthecirclex2+y2=4intersecttheparabolay=x2+k
Answered by 1549442205PVT last updated on 06/Aug/20
The circle intersection the parabol if and only if the system of equations { ((y=x^2 +k)),((x^2 +y^2 =4)) :} has the solution. Hence we need must to have : x^2 +(x^2 +k)^2 =4⇔x^4 +(2k+1)x^2 +k^2 −4=0(1) has the solution Set x^2 =t.(1)⇔t^2 +(2k+1)t+k^2 −4=0(2) The eqn.(1) has the solution if and only if the eqn.(2) has non−negative root which means equivalent to a)The case both roots are non−negative { ((Δ=(2k+1)^2 −4(k^2 −4)≥0)),((x_1 +x_2 =−(2k+1)≥0)),((x_1 x_2 =k^2 −4≥0)) :} ⇔ { ((4k+17≥0)),((k≤−(1/2))),((∣k∣≥2)) :}⇔ { ((k≥−((17)/4))),((k≤−(1/2))),((k∈(−∞;2]∪[2;+∞))) :} ⇔−((17)/4)≤k≤−(1/2) b)The case has only one root non−negative: { ((Δ=(2k+1)^2 −4(k^2 −4)≥0)),((x_1 x_2 =k^2 −4≤0)) :} ⇔ { ((4k+17≥0)),((∣k∣≤2)) :} ⇔ { ((k≥((−17)/4))),((−2≤k≤2)) :}⇔−2≤k≤2 Thus,The circle intersection the parabol if and only if k∈[−((17)/4);−(1/2)]∪[−2;2]
Thecircleintersectiontheparabolifandonlyifthesystemofequations{y=x2+kx2+y2=4hasthesolution.Henceweneedmusttohave:x2+(x2+k)2=4⇔x4+(2k+1)x2+k2−4=0(1)hasthesolutionSetx2=t.(1)⇔t2+(2k+1)t+k2−4=0(2)Theeqn.(1)hasthesolutionifandonlyiftheeqn.(2)hasnon−negativerootwhichmeansequivalenttoa)Thecasebothrootsarenon−negative{Δ=(2k+1)2−4(k2−4)⩾0x1+x2=−(2k+1)⩾0x1x2=k2−4⩾0⇔{4k+17⩾0k⩽−12∣k∣⩾2⇔{k⩾−174k⩽−12k∈(−∞;2]∪[2;+∞)⇔−174⩽k⩽−12b)Thecasehasonlyonerootnon−negative:{Δ=(2k+1)2−4(k2−4)⩾0x1x2=k2−4⩽0⇔{4k+17⩾0∣k∣⩽2⇔{k⩾−174−2⩽k⩽2⇔−2⩽k⩽2Thus,Thecircleintersectiontheparabolifandonlyifk∈[−174;−12]∪[−2;2]
Commented by bemath last updated on 05/Aug/20

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