Q41 / 110 · VITEEE 2024 · Mathematics
Given, $\frac{x^2+y^2}{x^2-y^2}+\frac{x^2-y^2}{x^2+y^2}=k$, then $\frac{x^8+y^8}{x^8-y^8}$ is equal to
- A$\frac{k^2+1}{k^2-1}$
- B$\frac{k^2+4}{k^2-4}$
- Correct answerC$\frac{k^2+4}{4 k}$
- D$\frac{k^2+8}{8 k}$
Solution
$\begin{aligned}
& \text { } \because \frac{x^2+y^2}{x^2-y^2}+\frac{x^2-y^2}{x^2+y^2}=k \\
& \Rightarrow \frac{\left(x^2+y^2\right)^2+\left(x^2-y^2\right)^2}{\left(x^2-y^2\right)\left(x^2+y^2\right)}=k \\
& \Rightarrow \quad \frac{2\left(x^4+y^4\right)}{\left(x^4-y^4\right)}=k \\
& \Rightarrow \quad \text { } \frac{x^4+y^4}{x^4-y^4}=\frac{k}{2} ...(i)\\
& \text { and } \\
& \frac{x^4-y^4}{x^4+y^4}=\frac{2}{k}...(ii)
\end{aligned}$
Adding Eqs. (i) and (ii),
$\begin{aligned}
& \Rightarrow \quad \frac{x^4+y^4}{x^4-y^4}+\frac{x^4-y^4}{x^4+y^4}=\frac{k}{2}+\frac{2}{k} \\
& \Rightarrow \frac{\left(x^4+y^4\right)^2+\left(x^4-y^4\right)^2}{\left(x^4+y^4\right)\left(x^4-y^4\right)}=\frac{k^2+4}{2 k} \\
& \Rightarrow \quad \frac{2\left(x^8+y^8\right)}{\left(x^8-y^8\right)}=\frac{k^2+4}{2 k}
\end{aligned}$
$\Rightarrow \quad \frac{x^8+y^8}{x^8-y^8}=\frac{k^2+4}{4 k}$
Hence, the correct answer is option: 'C'.