The orthocentre of the triangle $$PAB$$ 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 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
$${(x - 6)^2} + {y^2} = {(2\sqrt 3 )^2} \Rightarrow {x^2} + {y^2} - 12x + 24 = 0$$