Question

The length of the latusrectum of the ellipse 3x² + y² = 12 is
(a) 4
(b) 3
(c) 8
(d) $\frac{4}{\sqrt{3}}$

Answer 1

Given equation of ellipse is
$$\begin{gathered} 3x^2+y^2=12 \\ \mathrm{i}.\mathrm{e}\frac{3x^2}{12}+\frac{y^2}{12}=1 \\ \mathrm{i}.\mathrm{e}\frac{x^2}{4}+\frac{y^2}{{\left(\sqrt{12}\right)}^2}=1 \\ \mathrm{i}.\mathrm{e}\frac{x^2}{{\left(2\right)}^2}+\frac{y^2}{{\left(\sqrt{12}\right)}^2}=1 \\ \mathrm{since}a=2,b=\sqrt{12} \\ \mathrm{i}.\mathrm{e}b>a \\ \Rightarrow \mathrm{length}\mathrm{of}\mathrm{latus}\mathrm{rectum}\frac{2\times 4}{\sqrt{12}}=\frac{2\times 4}{2\sqrt{3}}=\frac{4}{\sqrt{3}} \end{gathered}$$
Hence, the correct answer is option D.