KEAM 2017 Math Paper

We have
Vertices and foci of hyperbola at (0,±15) and (0,±20)
since, both foci and vertices lies on Y -axis, then equation of hyperbola will be
‌

y2

b2

−‌

x2

a2

=1
Now, vertices =(0,±b)
∴‌‌b=15
Again, foci =(0,±be)
∴‌‌be=20
⇒‌‌e=‌

20

15

=‌

4

3

⇒‌‌√1+‌

a2

b2

=‌

4

3

  [∵e=√1+‌

a2

b2

]
⇒‌‌1+‌

a2

b2

=‌

16

9

⇒‌‌‌

a2

b2

=‌

7

9

⇒‌‌a2=‌

7

9

×b2=‌

7

9

×225
⇒‌‌a=‌

√7

3

×15=5√7
∴ Equation of hyperbola is
‌

y2

225

−‌

x2

175

=1