The slope of the tangent line to a plane curve y=f(x) at the point (x,y) is given by 3 - 2 m=- + 16x + 3 Vx Determine the equation of the curve which passes through the point (16 , 52) The equation of the curve is given by y = |

Equation is x^2+y^2 = 13, point (2,-3). Find the equation of the tangent OR normal at that point.

For the cruciform in the figure below, the equation is x2y2 - 9x² – 4y2 = 0, find the equation of the tangent line at the given point on the graph at (-4, 2V2) x-9x2-4y 0 4- (-4, 2/3)- +++ -6-4-2 2 46

please help!