Q. 224.8( 6 Votes )

Find the equation

Answer :


Let the equation of the required ellipse be


…(i)


Given:


Coordinates of foci = (±3, 0) …(ii)


We know that,


Coordinates of foci = (±c, 0) …(iii)


From eq. (ii) and (iii), we get


c = 3


We know that,


c2 = a2 – b2


(3)2 = a2 – b2


9 = a2 – b2


b2 = a2 – 9 …(iv)


Given that ellipse passing through the points (4, 1)


So, point (4, 1) will satisfy the eq. (i)


Taking point (4, 1) where x = 4 and y = 1


Putting the values in eq. (i), we get




[from (iv)]



16a2 – 144 + a2 = a2(a2 – 9)


17a2 – 144 = a4 – 9a2


a4 – 9a2 – 17a2 + 144 = 0


a4 – 26a2 + 144 = 0


a4 – 8a2 – 18a2 + 144 = 0


a2(a2 – 8) – 18(a2 – 8) = 0


(a2 – 8)(a2 – 18) = 0


a2 – 8 = 0 or a2 – 18 = 0


a2 = 8 or a2 = 18


If a2 = 8 then


b2 = 8– 9


= - 1


Since the square of a real number cannot be negative. So, this is not possible


If a2 = 18 then


b2 = 18 – 9


= 9


So, equation of ellipse if a2 = 18 and b2 = 9



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