The equation of tangent to the hyperbola `x^2/a^2-y^2/b^2=1` at `(a sec theta,b tan theta)` is `x/a sec theta-y/b tan theta=1`.

Given hyperbola is `x^2/a^2-y^2/b^2=1`.....................(1)

and the given line `y= mx +c` touches hyperbola then solve `(1) `and `(2)` , we get

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

or `(b^2 - a^2m^2) x^2 - 2a^2 mcx - a^2(c^2 + b^2) = 0`......................(3)

Since line `(2)` will be tangent to hyperbola `(1)`

if roots of equation `(3)` are equal i.e. `D = 0`

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

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

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

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

or `c^2 = a^2m^2- b^2` or `c= pm sqrt(a^2m^2-b^2)`. This is the required condition of tangency.

`y=mx pm sqrt(a^2m^2-b^2)` is a tangent to the standard hyperbola .............(1)

If above tangent passes through `(h, k)` then

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

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

Above equation is quadratic in m therefore it has two roots `m_1` and `m_2`.

Hence passing through a given point `(h, k)` there is a maximum of two tangents can be drawn to the hyperbola.

therefore `m_1+m_2=(2kh)/(h^2-a^2)`.........................(3)

`m_1m_2=(k^2+b^2)/(h^2-a^2)`..................................(4)

Equations `(3)` and `(4)` are used to find the locus of the point of intersection of a pair of tangents which enclose an angle `beta`.

Now `tan^2 beta= ((m_1+m_2)^2-4m_1m_2)/((1+m_1m_2)^2)`

(substituting the values of `m_1 + m_2` and `m_1m_2` to get the locus)

If `beta = 90^(circ)` then `m_1 m_2 =- 1` , hence from `(4)`

`k^2 + b^2 = a^2 - h^2`

`x^2 + y^2 = a^2 - b^2` which is the equation of director circle, of the given hyperbola.

The locus of the point of intersection of the tangents to the hyperbola `x^2/a^2-y^2/b^2=1`,

which are perpendicular to each other, is called the director circle.

Any tangent to hyperbola is `y = mx + sqrt(a^2m^2-b^2)`....................(1)

lf it passes through `P(h, k)`, then `k - mh = sqrt(a^2m^2-b^2)`

`k^2 + m^2h^2 - 2mk = a^2m^2 - b^2`

`m^2 (h^2- a^2) - 2mhk + (k^2 + b^2) = 0`

It is quadratic in `m` therefore it has two roots `m_1` and `m_2` . Hence two tangents (real or imaginary) can be drawn from `P(h, k)`.

If pair of perpendicular tangents are drawn from `P(h, k)` then `m_1 m_2 = (k^2+b^2)/(h^2-a^2)=-1`

`=> h^2+k^2 = a^2- b^2`

`:.` locus of `P(h, k)` is `x^2 + y^2 = a^2 - b^2`

Equation of normal to a hyperbola `x^2/a^2-b^2/b^2=1` at a point `(x_1,y_1)` is

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

The equation of normal at `(a sec theta, b tan theta)` to the hyperbola `x^2/a^2-y^2/b^2=1` is `i`

`ax cos theta + by cot theta = a^2 + b^2`

The combined equation of the pair of tangents drawn from a point `P(x_1 ,y_1)`, lying outside the hyperbola

`x^2/a^2-y^2/b^2 =1` is

`(x^2/a^2-y^2/b^2-1)(x_1^2/a^2-y_1^2/b^2-1)=((x x_1)/a^2-(yy_1)/b^2-1)^2` or `S S_1=T^2`

where `S=x^2/a^2-y^2/b^2-1 ; S_1=x_1^2/a^2-y_1^2/b^2-1` and `T=(x x_1)/a^2-(yy_1)/b^2-1`.