You need to find the derivative dy/dx of the curve and substitute x and y, if necessary, to find the slope at a particular given point. The slope is the tangent at the point. The standard linear equation y=ax+b can then be constructed, because the tangent will be a. To find b, the y intercept, substitute the coords of the given point into x and y and you'll be able to find b. That gives you the line.
For example, let y=px^2+qx+r be the equation of a parabola, where p, q and r are constants, then dy/dx=2px+q. Let's say you're given the point on the curve where x=A, then dy/dx=2pA+q. This is a number because p, q and A are all known values (you'll be told what they are). This is your slope for the tangent, so a=2pA+q, and y=(2pA+q)x+b is the equation of the tangent line. To find b we first find out what y is on the curve when x=A. We know y=pA^2+qA+r. To make it easier to read, call this y coord B. So we substitute (A,B) in our linear equation to find b=B-A(2pA+q). It's much easier when you have actual numbers rather than symbols. In the same way you can find the equation of the perpendicular because its slope is related to the slope of the tangent. It's -1/(2pA+q), the negative reciprocal of the tangent.
I hope this helps your understanding.