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Question Number 47480 by ajfour last updated on 10/Nov/18

Answered by MrW3 last updated on 10/Nov/18

assume the major and minor axes are parallel to x axis and y axis. eqn. (((x−h)^2 )/u^2 )+(((y−k)^2 )/v^2 )=1 h=((a+c)/2) k=((b+d)/2) (((a−h)^2 )/u^2 )+(((−k)^2 )/v^2 )=1 ...(i) (((−h)^2 )/u^2 )+(((b−k)^2 )/v^2 )=1 ...(ii) (((a−h)^2 (b−k)^2 −h^2 k^2 )/u^2 )=(b−k)^2 −k^2 (1/u^2 )=(((b−k)^2 −k^2 )/((a−h)^2 (b−k)^2 −h^2 k^2 ))=((b(b−2k))/([(a−h)(b−k)+hk][(a−h)(b−k)−hk])) ⇒(1/u^2 )=((4bd)/((ab+cd)(bc+ad))) (((b−k)^2 (a−h)^2 −h^2 k^2 )/v^2 )=(a−h)^2 −h^2 (1/v^2 )=(((a−h)^2 −h^2 )/((b−k)^2 (a−h)^2 −h^2 k^2 ))=((a(a−2h))/([(a−h)(b−k)+hk][(a−h)(b−k)−hk])) ⇒(1/v^2 )=((4ac)/((ab+cd)(bc+ad))) eqn. of ellipse: (((2x−a−c)^2 bd)/((ab+cd)(bc+ad)))+(((2y−b−d)^2 ac)/((ab+cd)(bc+ad)))=1 or ⇒(2x−a−c)^2 bd+(2y−b−d)^2 ac=(ab+cd)(bc+ad)

assumethemajorandminoraxesareparalleltoxaxisandyaxis.eqn.(x−h)2u2+(y−k)2v2=1h=a+c2k=b+d2(a−h)2u2+(−k)2v2=1...(i)(−h)2u2+(b−k)2v2=1...(ii)(a−h)2(b−k)2−h2k2u2=(b−k)2−k21u2=(b−k)2−k2(a−h)2(b−k)2−h2k2=b(b−2k)[(a−h)(b−k)+hk][(a−h)(b−k)−hk]⇒1u2=4bd(ab+cd)(bc+ad)(b−k)2(a−h)2−h2k2v2=(a−h)2−h21v2=(a−h)2−h2(b−k)2(a−h)2−h2k2=a(a−2h)[(a−h)(b−k)+hk][(a−h)(b−k)−hk]⇒1v2=4ac(ab+cd)(bc+ad)eqn.ofellipse:(2x−a−c)2bd(ab+cd)(bc+ad)+(2y−b−d)2ac(ab+cd)(bc+ad)=1or⇒(2x−a−c)2bd+(2y−b−d)2ac=(ab+cd)(bc+ad)

Commented by ajfour last updated on 10/Nov/18

Thank you Sir. (Matches!)

ThankyouSir.(Matches!)

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