Question

A hyperbola touches y axis and has its centre at (5/2,20)and one of the focii at (10/24) respectively , find length of the transverse axis

Answer 1

Logic Used :
1)Foot of Perpendicular from focus on any tangent to the Hyperbola will lie on auxiliary circle.
2) Distance between two points.

I didn't shown graph of Hyperbola .

Steps:
1) Centre of Hyperbola ,B : (2.5,20)
One of its Focus ,A: (10,24)

Since, Hyperbola touches Y-axis.
=> y-axis is it's Tangent.

So, Clearly
Foot of Perpendicular of Focus on Tangent (y-axis) : C = (0,24)

2) Equation of Auxiliary Circle :
${(x - 2.5)}^{2} + {(y - 20)}^{2} = {a}^{2}$

where :
2a => Length of Transverse Axis.

3) C lies on this circle :

=> BC = a
$= > a = \sqrt{ {(2.5 - 0)}^{2} + {(20-24)}^{2} } \\ = > a = \sqrt{ {2.5}^{2} + {4}^{2} } \\ = > a = \sqrt{22.25} \: units$
Now,
Length of Transverse axis :
$2a = \sqrt{4 \times {a}^{2} } \\ = > \sqrt{4 \times 22.25} = \sqrt{89} \: units$

So, Desired Length of Transverse Axis
: √89 units