Tangents are drawn from the point $$P(3, 4)$$ to the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 4} = 1$$ touching the ellipse at points $$A$$ and $$B$$.
The equation of the locus of the point whose distances from the point $$P$$ and the line $$AB$$ are equal, is
Tangents are drawn from the point $$P(3, 4)$$ to the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 4} = 1$$ touching the ellipse at points $$A$$ and $$B$$.
The coordinates of $$A$$ and $$B$$ are
Equation of the circle with $$AB$$ as its diameter is
A point on hyperbola is (3sec$$\theta$$, 2tan$$\theta$$).
It lies on the circle, so
$$9{\sec ^2}\theta + 4{\tan ^2}\theta - 24\sec \theta = 0$$.
$$ \Rightarrow 13{\sec ^2}\theta - 24\sec \theta - 4 = 0 \Rightarrow \sec \theta = 2, - {2 \over {13}}$$
Therefore, $$\sec \theta = 2 \Rightarrow \tan \theta = \sqrt 3 $$
The point of intersection are $$A(6,2\sqrt 3 )$$ and $$B(6, - 2\sqrt 3 )$$
Hence, The circle with AB as diameter is
$${(x - 6)^2} + {y^2} = {(2\sqrt 3 )^2} \Rightarrow {x^2} + {y^2} - 12x + 24 = 0$$
Equation of a common tangent with positive slope to the circle as well as to the hyperbola is
A tangent to $${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$$ is
$$y = mx + \sqrt {9{m^2} - 4} ,\,m > 0$$ .... (1)
A tangent to $${(x - 4)^2} + {y^2} = 16$$ is
$$xy = m(x - 4) + 4\sqrt {1 + {m^2}} $$ ..... (2)
Comparing (1) and (2),
$$\sqrt {9{m^2} - 4} = - 4m + 4\sqrt {1 + {m^2}} \Rightarrow \sqrt {9 - {4 \over {{m^2}}}} = - 4 + 4\sqrt {1 + {1 \over {{m^2}}}} $$
Let $${1 \over {{m^2}}} = t$$, we have $$\sqrt {9 - 4t} = - 4 + 4\sqrt {1 + t} $$
Squaring, we have
$$ \Rightarrow 9 - 4t = 16 + 16(1 + t) - 32\sqrt {1 + t} \Rightarrow 32\sqrt {1 + t} = 23 + 20t$$
Again squaring $$1024(1 + t) = 529 + 920t + 400{t^2}$$
$$ \Rightarrow 400{t^2} - 104t - 495 = 0 \Rightarrow t = {5 \over 4}$$
Thus $${m^2} = {4 \over 5},\,m = {2 \over {\sqrt 5 }}$$
The tangent is $$y = {2 \over {\sqrt 5 }}x + {4 \over {\sqrt 5 }}$$ i.e. $$2x - \sqrt 5 y + 4 = 0$$