Question -
Answer -
Given:
The point (-3, 1)
Eccentricity = тИЪ(2/5)
Now let us find the equation to the ellipse.
We know that the equation of the ellipse whose axes are xand y тАУ axis is given as
тАж. (1)
Now let us substitute equation (2) in equation (1), we get
It is given that the curve passes through the point (-3, 1).
So by substituting the point in the curve we get,
3(- 3)2┬а+ 5(1)2┬а= 3a2
3(9) + 5 = 3a2
32 = 3a2
a2┬а= 32/3
From equation (2)
b2┬а= 3a2/5
= 3(32/3) / 5
= 32/5
So now, the equation of the ellipse is given as:
3x2┬а+ 5y2┬а= 32
тИ┤┬аThe equation of the ellipse is 3x2┬а+5y2┬а= 32.