about to only mathematics

Answer

(0, 2)

Explanation

Let ` P ( a cos theta, 2 sin theta ) ` be a point on the ellipse
` 4 x ^(2) + a ^(2) y ^(2)= 4 a ^(2), i.e. ( x^(2))/( a^(2)) + ( y ^(2))/( 4) =1 `
Let ` A (0, -2)` be the given point.
Then,
` (AP) ^(2) = a ^(2) cos ^(2) theta + 4 (1 + sin theta ) ^(2)`
`rArr (d) /( d theta ) ( AP ) ^(2) = - a ^(2) sin 2 theta + 8 ( 1 + sin theta ) * cos theta `
` rArr (d)/( d theta ) (AP ) ^(2) = [ (8 - 2a ^(2)) sin theta + 8 ] cos theta `
For maximum or minimum, we put `(d)/(d theta ) (AP) ^(2) = 0 `
`rArr [ ( 8- 2a ^(2)) sin theta + 8 ] cos theta = 0`
`rArr cos theta = 0 or sin theta = ( 4)/(a ^(2) - 4)`
`[ because 4lt a ^(2) lt 8 rArr (4)/(a ^(2) - 4) gt 1 rArr sin theta gt 1`, which is impossible]
Now, ` (d ^(2) ) /( d theta ^(2)) (AP) ^(2) = - {( 8 - 2 a ^(2)) sin theta + 8} sin theta + ( 8 - 2a^(2) ) * cos ^(2) theta `
For ` theta = ( pi )/(6)`, we have ` (d ^(2))/(d theta ^(2)) (AP ) ^(2) = - ( 16 - 2a ^(2)) lt 0`
Thus, ` AP^(2)` i.e. AP is maximum when `theta = (pi)/(2)`. The point on the curve `4 x ^(2) + a ^(2) y ^(2) = 4a ^(2) ` that is farthest from the point
`A( 0 , - 2)` is ` ( a cos ""(pi)/(2), 2 sin ""(pi)/(2))= (0, 2)`