JEE Main 10 April 2015 Paper

x2

4

−

y2

9

=1
focii are (√13,0) and (−√13,0)
eccentricity of hyperbola is eH=

√13

2

  ⇒ eccentricity of ellipse is eE=

1

√13

Let equation ellipse is

x2

a2

+

y2

b2

=1    ⇒eE2=1−

b2

a2

⇒

1

13

=1−

b2

a2

           ⇒

b

a

=√

12

13

..(1)
Ellipse passes through (±√13,0)    ⇒

13

a2

=1
⇒a2=13 ⇒b=√12
Equation of ellipse =

x2

13

+

y2

12

=1
which is passes through (√13,0)(√

13

2

,√6) and (

√39

2

,√3)