The length of the latus rectum of $16 x^2+25 y^2=400$ is

The length of the latus rectum of $16 x^2+25 y^2=400$ is
  1. $\frac{25}{2}$
  2. $\frac{25}{4}$
  3. $\frac{16}{2}$
  4. $\frac{32}{5}$

Solution

$\begin{aligned} & 16 x^2+25 y^2=400 \Rightarrow \frac{x^2}{25}+\frac{y^2}{16}=1 \\ a^2= & 25, b^2=16 \end{aligned}$
Length of lectus rectum $=\frac{2 b^2}{a}=\frac{32}{5}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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