AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

Consider the equation of the hyperbola be ‌

x2

a2

−‌

y2

b2

=1
It is given that the centre of the hyperbola is (0,0) and its
transverse axis is along x-axis and its length is 12
2a=12
a=6
Substitute this value in equation of hyperbola.
‌

x2

62

−‌

y2

b2

=1
The above equation is passes through point (8,2)
‌

82

62

−‌

22

b2

=1
‌

64

36

−‌

4

b2

=1
After simplification,
b=‌

6

√7

The equation of hyperbola becomes,
‌

x2

62

−‌

y2

(‌ 6 √7 )2

=1
‌

x2

36

−‌

7y2

36

=1
The eccentricity of hyperbola is,
e=√1+‌

a2

b2

e=√1+‌

62

( 6 √7 )2

e=√‌

8

7

e=‌

2√2

√7