JEE Mains 03-Sep-2020 Shift 1 Solved Paper

Ellipse :

x2

4

+

y2

3

=1
eccentricity =√1−

3

4

=

1

2

∴ foci =(±1,0)
for hyperbola, given 2 a=√2⇒a=

1

√2

∴ hyperbola will be

x2

1∕2

−

y2

b2

=1
eccentricity =√1+2b2
∴ foci =(±√

1+2b2

2

,0)
Ellipse and hyperbola have same foci
⇒ √

1+2b2

2

=1
⇒ b2=

1

2

∴ Equation of hyperbola :

x2

1∕2

−

y2

1∕2

=1
⇒ x2−y2=

1

2

Clearly (√

3

2

,

1

√2

) does not lie on it.