Q.Find the point on the curve $y = \sqrt{x-3}$ where the tangent is perpendicular to the line $6x + 3y - 5 = 0$.

The derivative dy/dx evaluated at a point $(x_1,y_1)$ on a curve $y=f(x)$ gives the slope $m$ of the tangent line to the curve at that point — this is the geometric meaning of the derivative. The tangent is the straight line that just touches the curve at $(x_1,y_1)$ and has the same instantaneous direction as the curve there; its equation is $y-y_1=m(x-x_1)$. The normal is the line through the same point that is perpendicular to the tangent; since perpendicular lines have slopes that are negative reciprocals of each other, the normal's slope is $m'=-1/m$ (valid when $m\ne0$), and its equation is $y-y_1=m'(x-x_1)$. When the curve is given implicitly (an equation in $x$ and $y$ that cannot easily be solved for $y$) or parametrically (both $x$ and $y$ given in terms of a parameter such as $\theta$ or $t$), the same idea applies: differentiate implicitly using the chain and product rules, or use $dy/dx=(dy/d\theta)/(dx/d\theta)$ for parametric curves, evaluate the resulting slope at the given point, and substitute into the point-slope forms above. Special cases: a horizontal tangent (slope $0$) gives a vertical normal, and a vertical …