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

$=>a=\sqrt{(2.5-0)^2+(20-24)^2}\\ =>a=\sqrt{22.25}$

Now,

length of transverse axis :

$2a=\sqrt{4*a^2}\\    = \sqrt{89}$

So, Desired length of Transverse axis : $\sqrt{89}$