Here's (b)
x^2 + xy - y^2 = 1 using implicit differentiation, we have
2x + y + xy' - 2yy' = 0
xy' - 2yy' = -2x - y multiply through by - 1 on both sides
2yy' - xy' = 2x + y
y'(2y - x) = 2x + y
y' = [2x + y ] / [2y - x ] and the slope of the tangent line at (2.3) is given by
y' = [2(2) + 3] / [2(3) - 2 ] = [7] / [ 6 - 2] = 7 / 4
Here's a graph of the tangent line to the "rotated" hyperbola at the point (2,3)....https://www.desmos.com/calculator/yuxxmj7ud9
Note, that the slope of the tangent line at this point = (7/4).....just as we expected...!!!
Here's (b)
x^2 + xy - y^2 = 1 using implicit differentiation, we have
2x + y + xy' - 2yy' = 0
xy' - 2yy' = -2x - y multiply through by - 1 on both sides
2yy' - xy' = 2x + y
y'(2y - x) = 2x + y
y' = [2x + y ] / [2y - x ] and the slope of the tangent line at (2.3) is given by
y' = [2(2) + 3] / [2(3) - 2 ] = [7] / [ 6 - 2] = 7 / 4
Here's a graph of the tangent line to the "rotated" hyperbola at the point (2,3)....https://www.desmos.com/calculator/yuxxmj7ud9
Note, that the slope of the tangent line at this point = (7/4).....just as we expected...!!!