Find the point (x, y) of x^2+14xy+49y^2=100 that is closest to the origin and lies int he first quadrant

The given equation may be simplified as follows:

x² + 14xy + 49y² = 100
(x + 7y)(x + 7y) = 100
(x + 7y)² = 10²
x + 7y = 10

This is a straight line with the equation
y = -(1/7)x + 10/7

The minimum distance from the origin to this line is provided by a straight line that passes through the origin and which is perpendicular to the straight line.

The slope of the perpendicular line is 7 because the product of the two slopes should be -1.
The perpendicular line is of the form
y = 7x + c.

Because the line passes through (0,0), therefore c = 0.
The line y = 7x intercepts the original line when
y = 7x = -(1/7)x + 10/7

Therefore
7x = -(1/7)x + 10/7
Multiply through by 7.
49x = -x + 10
50x = 10
x = 1/5
y = 7x = 7/5

The minimum distance is
d = √(x² + y²)
   = √[(1/5)² + (7/5)²]
   = √2

The point is (1/5, 7/5).

(1/5, 7/5)  or (0.2, 1.4)
 

The answer to your question is b’(16,14)