Find the equation of the line tangent to the curve y=(x^2+3)^1/2 that is perpendicular to the line 2x-y+7=0

Answer 1

you want the tangent to have slope -1/2

y' = x/√(x^2+3)

so, where is that = -1/2 ?

Get the point and use the point-slope form of the line.