Question

The distance between the foci of the ellipse $3x^2+4y^2=48$ is

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (B)

Given:

Equation of the ellipse is $3x^2+4y^2=48$

$\Rightarrow \frac{3x^2}{48}+\frac{4y^2}{48}=1$

$\Rightarrow \frac{x^2}{16}+\frac{y^2}{12}=1$

Here, $16>12$

Lets compare with standard form of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b$

$a^2=16$ and $a=4$---(1)

$b^2=12$

$e=\sqrt{1-\frac{b^2}{a^2}}$

$\Rightarrow e=\sqrt{1-\frac{12}{16}}$

$\Rightarrow e=\sqrt{\frac{16-12}{16}}$

$\Rightarrow e=\sqrt{\frac{4}{16}}$

$\Rightarrow e=\frac{2}{4}$

$\Rightarrow e=\frac{1}{2}$ --(2)

Since, $a>b$, Major axis is $a$.

$\Rightarrow$ Distance between the foci $=2ae$

$=2\times 4\times \frac{1}{2}$

$=4$

Therefore, Distance between the foci is $4$.