Define : `tan phi =dy/dx|_(P)`

(1) Equation of a tangent at `P (x_1, y_1)`
`y-y_1=dy/dx|_(x_1,y_1) (x-x_1)`

(2) Equation of normal at `(x_1, y_1)`

`y-y_1 = -(1/ (dy/dx)_(x_1,y_1) ) (x -x_1)`, if `dy/dx ]_(x_1,y_1)` exists.

However in some cases `dy/dx` fails to exist but still a tangent can be drawn e.g. case of vertical tangent.

Note that the point `(x_1, y_1)` must lie on the curve for the equation of tangent and normal.

Important notes to remember:
(a) If `dy/dx|_(x_1,y_1) =0 => ` tangent is parallel to `x`-axis and converse.

If tangent is parallel to `ax + by+ c = 0 => dy/dx =-a/b`

(b) If `dy/dx|_(x_1,y_1) ->oo` or `dx/dy|_(x_1,y_1) =0 => ` tangent is perpendicular to `x`-axis.

If tangent with a finite slope is perpendicular to `ax + by+ c = 0`

`=> dy/dx|_(x_1,y_1) * (-a/b) =-1`

(c) If the tangent at `P (x_1, y_1)` on the curve is equally inclined
to the coordinate axes

`=> dy/dx|_(x_1,y_1) = pm 1`.

(d) If the tangent makes equal non zero intercept on
the coordinate axes then `dy/dx|_(x_1,y_1) =-1`

(e) If tangent cuts off from the coordinate axes equal distance om the origin `=>dy/dx= pm1`


(f) `OT` is called the initial ordinate of the tangent

`Y-y =dy/dx (X-x)`

put `X=0` to get

`:.` `Y=OT=y-x dy/dx` (It is the `y` intercept of a tangent at `P`)

(g) Concept: `F(x)= f(x)*g(x)` are such that `f(x)` is continuous at `x = a` and `g(x)` is differentiable at `x = a` with
`g(a)=0` then the product function `f(x)*g(x)` is diffenentiable at `x = a`.