Menu Close

Find-the-symmetry-axes-if-any-of-the-following-curve-x-2-y-2-2xy-2x-8y-2-0-

Tinku Tara 0 Comments
Pin




Question Number 85298 by mr W last updated on 20/Mar/20
Find the symmetry axes (if any) of the following curve: x^2 +y^2 −2xy+2x−8y−2=0
Findthesymmetryaxes(ifany)ofthefollowingcurve:x2+y2−2xy+2x−8y−2=0
Commented by jagoll last updated on 20/Mar/20
x^2 +2x+1+y^2 −8y+16 = 2xy+19 (x+1)^2 +(y−4)^2 = 2xy+19 next
x2+2x+1+y2−8y+16=2xy+19(x+1)2+(y−4)2=2xy+19next
Commented by mr W last updated on 20/Mar/20
any result sir?
anyresultsir?
Answered by MJS last updated on 21/Mar/20
y=x+4±(√(6(x+3))) defined for x≥−3 x=y−1±(√(3(2y+1))) defined for y≥−(1/2) ⇒ it′s a parabola with direction x^+ and y^+ ...under construction...
y=x+4±6(x+3)definedforx⩾−3x=y−1±3(2y+1)definedfory⩾−12⇒it′saparabolawithdirectionx+andy+…underconstruction…
Commented by mr W last updated on 22/Mar/20
thank you sir! i got the symmetry axis y=x+(5/2) and have checked graphically.
thankyousir!igotthesymmetryaxisy=x+52andhavecheckedgraphically.
Commented by mr W last updated on 21/Mar/20
Answered by mr W last updated on 21/Mar/20
Method 1 x^2 +y^2 −2xy+2x−8y−2=0 ⇒(x−y+1)^2 −6y−3=0 it is to see that the curve is a parabola, which has one and only one symmetry axis. the symmetry axis of a parabola is one of the lines which intersect (not tangent) the parabola at only one point. let′s say this symmetry axis is x=my+k, or y=(x/m)−(k/m). the intersection of both is the solution of { (((x−y+1)^2 −6y−3=0)),((x=my+k)) :} (my+k−y+1)^2 −6y−3=0 [(m−1)y+(k+1)]^2 −6y−3=0 (m−1)^2 y^2 +2[(m−1)(k+1)−3]y+(k+1)^2 −3=0 since it should always have only one root, the y^2 term must vanish, i.e. m=1. the eqn. is then −6y+(k+1)^2 −3=0 ⇒y=(((k+1)^2 −3)/6) ⇒x=(((k+1)^2 −3)/6)+k such that this line is parabola′s axis, the intersection point is the vertex, i.e. vertex is ((((k+1)^2 −3)/6)+k, (((k+1)^2 −3)/6)). we know at vertex the tangent of the parabola is perpendicular to the (symmetry) axis of the parabola, i.e. y′=−m=−1. from (x−y+1)^2 −6y−3=0 we get 2(x−y+1)(1−y′)−6y′=0 at the vertex: 2(k+1)(1+1)+6=0 ⇒k=−(5/2) therefore the symmetry axis is x=my+k=y−(5/2) or y=x+(5/2)
Method1x2+y2−2xy+2x−8y−2=0⇒(x−y+1)2−6y−3=0itistoseethatthecurveisaparabola,whichhasoneandonlyonesymmetryaxis.thesymmetryaxisofaparabolaisoneofthelineswhichintersect(nottangent)theparabolaatonlyonepoint.let′ssaythissymmetryaxisisx=my+k,ory=xm−km.theintersectionofbothisthesolutionof{(x−y+1)2−6y−3=0x=my+k(my+k−y+1)2−6y−3=0[(m−1)y+(k+1)]2−6y−3=0(m−1)2y2+2[(m−1)(k+1)−3]y+(k+1)2−3=0sinceitshouldalwayshaveonlyoneroot,they2termmustvanish,i.e.m=1.theeqn.isthen−6y+(k+1)2−3=0⇒y=(k+1)2−36⇒x=(k+1)2−36+ksuchthatthislineisparabola′saxis,theintersectionpointisthevertex,i.e.vertexis((k+1)2−36+k,(k+1)2−36).weknowatvertexthetangentoftheparabolaisperpendiculartothe(symmetry)axisoftheparabola,i.e.y′=−m=−1.from(x−y+1)2−6y−3=0weget2(x−y+1)(1−y′)−6y′=0atthevertex:2(k+1)(1+1)+6=0⇒k=−52thereforethesymmetryaxisisx=my+k=y−52ory=x+52

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *