Let the given line is `y = mx + c`...............................(1) and given ellipse is `x^2/a^2+y^2/b^2=1`......................(2)

If line touch is the ellipse then by solving the two equations simultaneously (by eliminating `y` from `(I)` & `(2)`, we get

`x^2/a^2 + ((mx+c)^2)/b^2=1`

i.e. `(b^2 + a^2m^2) x^2 + 2a^2 cmx + a^2 (c^2- b^2) = 0` .........................(3)

Since line is tangent to the ellipse therefore its `D = 0`

`4a^4c^2m^2- 4 (b^2 + a^2m^2) * a^2( c^2 - b^2) = 0`

or `4a^2 [ a^2c^2m^2 - b^2c^2 - a^2c^2m^2 + b^4 + a^2b^2m^2] = 0`

or `b^2 (- c^2 + b^2 + a^2m^2) = 0`

or `c^2=b^2+a^2m^2` or `c= pm sqrt(a^2m^2+b^2)`

which is the required condition of tangency.

Substituting this value of `c` in `y = mx +c`, we have

`y=mx+sqrt(a^2m^2+b^2)` or `y=mx -sqrt(a^2m^2+b^2)` , which are tangents to the ellipse for all values of `m`.

Here `pm` sign represents two tangents to the ellipse having the same `m`, i.e. there are two tangents parallel to any given direction.

The equation of any tangent to the ellipse `((x-h)^2)/a^2+((y-k)^2)/b^2=1`

`(y-k)=m(x-h) pm sqrt (a^2m^2+b^2)`

`(xcos theta)/a+ (y sin theta)/b=1` is tangent to the ellipse at the point `(a cos theta, b sin theta)`

`(i)` Point of intersection of the tangents at the point `alpha` & `beta` is

`(a(cos (alpha+beta)/2)/(cos (alpha-beta)/2), b(sin (alpha+beta)/2)/(cos (alpha-beta)/2))` can be deduced by

comparing chord joining `P(alpha)` and `Q(beta)` with C.O.C. of the pair of tangents from `R(x_1, y_1)` on the ellipse,

`x_1= a(cos (alpha+beta)/2)/(cos (alpha-beta)/2); y_1=b(sin (alpha+beta)/2)/(cos (alpha-beta)/2)`

`(ii)` The eccentric angles of point of contact of two parallel tangents differ by `pi`. Conversely if the difference between the eccentric angles of two points is `pi` then the tangents at these points are parallel.

Equation of the tangent to the ellipse at `(x_1,y_1)` is `(x x_1)/a^2+(y y_1)/b^2=1.`

The slope of the tangent at `(x_1,y_1)=(-x_1)/(a^2) xx b^2/y_1`

`:. ` Slope of the normals at `(x_1,y_1)=a^2/x_1 xx y_1/b^2=(a^2y_1)/(b^2x_1)`

Hence the equation of the normal at `(x_1 ,y_1)` is `y - y_1=(a^2y_1)/(b^2x_1)(x-x_1)`

or `(x-x_1)/(x_1/a^2)=(y-y_1)/(y_1/b^2)`

ln above equation if we put `x = a cos theta` and `y= b sin theta` then we will get normal equation in parametric form.

`:. ax sec theta - b y cosec theta = a^2 - b^2 = a^2e^2`

This is equation of normal in parametric form.

Equation of a normal in terms of its slope `'m'` is `y = mx-((a^2-b^2)m)/(sqrt(a^2+b^2m^2))`

Locus of the point of intersection of the tangents which meet at right angles is called the Director Circle.

Let equation of any tangent is `y= mx + sqrt(a^2m^2 + b^2)`

If it passes through `(h, k)` then

`k =mh pm sqrt (a^2m^2+b^2)`

`(k - mh)^2 = a^2m^2 + b^2`

`(h^2- a^2)m^2- 2khm + k^2- b^2 = 0` .... (3)

Equation `(3)` has two roots `m_1` & `m_2`

`m_1+m_2=(2hk)/(h^2-a^2)` ................(4)

`m_1m_2=(k^2-b^2)/(h^2-a^2)` ..............(5)

Hence passing through a given point there can be a maximum of two tangents.

If `PA bot PB` then `m_1m_2=-1`

i.e., `m_1m_2=(k^2-b^2)/(h^2-a^2)=-1`

i.e., `k^2-b^2=a^2-h^2` ;

i.e., `x^2+y^2=a^2+b^2`

which is the director circle of the ellipse. Hence director circle of an ellipse is a circle whose centre is the centre of ellipse and whose radius is the length of the line joining the ends of the major and minor axis.

Equation `(3)` can be used to determine the locus of the point of intersection of two tangents enclosing.

If from any point `P (h, k)` pair of tangents are drawn to the ellipse which include an angle a , then

`tan alpha= |(m_1-m_2)/(1+m_1m_2)|=(sqrt((m_1+m_2)^2-4m_1m_2))/(1+m_1m_2)`

By putting value of `m_1 + m_2` and `m_1m_2` in above equation we will get the angle between pair of tangents.

Pair of tangents `PA` and `PB` are drawn from outside point `P (x_1, y_1)`, which is shown below. Hence joint equation of line `PA` and `PB` is given by `SS_1 = T^2`

Here `S equiv x^2/a^2+y^2/b^2-1`

`S_1 equiv x_1^2/a^2+y_1^2/b^2-1`

`T equiv (x x_1)/a^2+(yy_1)/b^2-1`