
Thanks for pointing out the error in solution of 4th one Gaurika. Here it is again for you in .... two different ways.
4. x^{2} + 9y^{2}  4x + 6y + 4 = 0 i.e. {x^{2}  2(x)(2) + 2^{2}} + {(3y)^{2} + 2(3y)(1) + 1} = 1 i.e. (x  2)^{2} + (3y + 1)^{2} = 1
Now we want to maximise 4x  9y. Let 4x  9y = k, i.e. x = (k + 9y)/4
So eliminating x, above equation becomes: (k + 9y  8)^{2} + 16(3y + 1)^{2} = 16 i.e. 225y^{2} + {18(k  8) + 96}y + (k  8)^{2} = 0 i.e. 225y^{2} + (18k  48)y + (k  8)^{2} = 0
Now as y is a real number, so discriminant of the above equation should be greater than or equal to zero. So, (18k  48)^{2}  4(225)(k  8)^{2} ≥ 0 i.e. (18k  48)^{2}  (30k  240)^{2} ≥ 0 i.e. (3k  8)^{2}  (5k  40)^{2} ≥ 0 i.e. (2k + 32)(8k  48) ≥ 0 i.e. (k  6)(k  16) ≤ 0 i.e. 6 ≤ k ≤ 16.
Thus required maximum value of 'k' is 16.
Alternate approach is geometric one We know that (x  2)^{2} + (3y + 1)^{2} = 1 and we need to maximise 4x  9y.
Now, let's replace 3y by Y, so that know equation turns into a circle (x  2)^{2} + (Y + 1)^{2} = 1 with center at (2, 1) and radius 1. And now we need to find the maximum value of 4x  3Y which will be equal to 4x  3Y = k (say) and will be achieved when this line is tangential to the circle.
Rest part is simple. I hope you know how to find the perpendicular distance of a point from a line.
Using the relation for the distance, we get that {4(2)  3(1)  k}/√(4^{2} + 3^{3}) = 1 i.e. 11  k = 5 i.e. k = 6 or 16.
Thus maximum value of k = 16 as found earlier too.
Kamal Lohia
