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Question

Find the equation of the ellipse satisfying the given condition e=34, foci on Y-axis, centre at origin and passes through (6,4).

Or
Find the equation of the hyperbola with vertices at (±5,0) and foci (±7,0)

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Solution

Let the equation of ellipse be x2a2+y2b2=1 (i)

If the ellipse passes through (6,4) then
(6)2a2+(4)2b2=136a2+16b2=11 (ii)

We know that, e=1a2b2 [b>a]

34=1a2b2 916=1a2b2

a2b2=1916=716 a2=716b21 (iii)

From Eqs. (i) and (iii), we get

36×167b2+16b2=1 167b2(36+7)=1

1b2=716×43 b2=16×437

Put the value of b^2 in Eq. (iii) we get

a2=716×16×437=43

On substituting the values of a2 and b2 in Eq. (i), we get

x243+7y216×43=1

16x2+7y2=688, which is the required equation of ellipse.

Or

Since, the vertices are on the X-axis with origin at the mid-point.
(±a,0)=(±5,0)and(±c,0)=(±7,0)

Then, equation of the hyperbola is of the form
x2a2y2b2=1

Since, the vertices are (±5,0) and foci are (±7,0])
a=5 and c=7

We know that, in hyperbola
c2=a2+b2

72=52+b2b2=4925=24

On putting the value of b2 in Eq. (i) we get
x225y224=124x225y2=600


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