JEE Main Math Class 11 Coordinate Geometry Part 1 Questions

Equation of hyperbola is
‌

x2

4

−‌

y2

9

=1
Its Foci =(±√13,0) e=‌

√13

2

If e, be the eccentricity of the ellipse, then
e1×‌

√13

2

=‌

1

2

⇒e1=‌

1

√13

Equation of ellipse is
‌

x2

a2

+‌

y2

b2

=1
Since ellipse passes through the foci (±√13,0) of the hyperbola, therefore
a2=13
Now √a2−b2=ae1
∴13−b2=1
⇒b2=12
Hence, equation of ellipse is
‌

x2

13

+‌

y2

12

=1
Now putting the coordinate of the point (‌

√13

2

,‌

√3

2

) in the equation of the ellipse, we get
‌

13

4×13

+‌

3

4×12

=1
⇒‌

1

4

+‌

1

16

=1, which is not true,