Using the two points, I managed to get the equation of the transverse axis as: $x – 3y + 14 = 0$
Conjugate axis as: $3x + y – 8 = 0$
Centre = $(1,5)$
$2ae =$ Distance between the foci = $ \sqrt{40} $
But now, to get the equation of the Hyperbola, I can use the definition of hyperbola which says
$($Distance from conjugate axis$)^2/a^2 – ($Distance from transverse axis$)^2/b^2 =1$
But I am having trouble finding the '$a$' and '$b$' values. I know I need to make use of the condition that $y$ axis is one of the tangents
Best Answer
The locus of intersections of tangents to a hyperbola with perpendiculars drawn from each focus is a circle with radius equal to the semimajor axis.
So, you can take either $A_1(0,4)$ or $A_2(0,6)$ as the intersection point and compute its distance from the center $C$ to get $a$, then use $f^2=a^2+b^2$ to find $b$.
In the same vein, all points symmetric to a focus with respect to tangents to the hyperbola lie on a circle with radius equal to twice the semimajor axis centered on the other focus:
Therefore, you can compute $2a$ by either taking $B_1(2,4)$ and finding its distance from $(4,6)$ or taking $B_2(-4,6)$ and finding its distance from $(-2,4)$. From there you can again recover $b$ via the equation $f^2=a^2+b^2$.
With the semiaxis lengths, center and transverse axis in hand, producing an equation for the hyperbola should be a simple matter. Note that for this equation, you really need $a^2$ and $b^2$ and not $a$ and $b$ themselves, so you can avoid taking any square roots in the above calculations.