TS EAMCET 19-July-2022 Shift 2 Solved Paper

Let the equation of hyperbola whose centre is at origin and transverse axis is along X-axis is
‌

x2

a2

−‌

y2

b2

=1
As the length of transverse axis is 8,2a=8
⇒‌‌a=4
The equation of hyperbola is ‌

x2

16

−‌

y2

b2

=1
The hyperbola passes through (5,2)
‌

25

16

−‌

4

b2

=1
⇒‌‌‌

4

b2

=‌

25

16

−1⇒b2=‌

64

9

∴ The equation of hyperbola is
‌

x2

16

−‌

y2

64∕9

=1
Equation of conjugate hyperbola is
‌

x2

16

−‌

y2

64∕9

=−1
‌ Eccentricity ‌‌=√1+‌

a2

b2

=√1+‌

16

64∕9

‌=√1+‌

16×9

64

=√1+‌

9

4

=‌

√13

2