# Find the equation 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 Rate this question :

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