The equation of the tangent to the curve `x^2+2y=8` which is the perpendicular to `x-2y+1=0` is

The equation of the tangent to the curve x^2+2y=8 and which is perpendicular to x-2y+1=0 is

The equation of the tangent to the curve y=x^3+3x^2-5 and which is perpendicular to 2x-6y+1=0 is

Find the equation of the tangent to the curve. y=x^(3)+3x^(2)-5 and which is perpendicular to 2x-6y+1=0

The equation of the tangent to the curve y=8/(4+x^2) at x=2 is

Find the equations of the tangents to the curve 3x^2 - y^2 = 2 which are perpendicular to the line x + 3y = 2.