Question

The distance between the foci of the ellipse \[3x^2+4y^2=48\] is______.

पर्याय

• 2

• 4

• 6

• 8

Answer 1

The distance between the foci of the ellipse \[3x^2+4y^2=48\] is 4.

Explanation:

The given ellipse is: \[3x^2+4y^2=48\]

Dividing both sides by 48:

                                     \[\frac{x^2}{16}+\frac{y^2}{12}=1\]

Here:

• \[a^2=16\] (major axis along x-axis)
• \[b^2=12\]

Find \( c \):

             \[c^2=a^2-b^2=16-12=4\Rightarrow c=2\]

The two foci are at \[(\pm c,0)=(\pm2,0).\]

Distance between the foci:

                                      \[=2c=2\times2=4\]