Question

If $\frac{x^2+y^2}{x^2-y^2}=\frac{17}{8}$, then $\frac{x}{y}$?

• A. $\frac{6}{8}$
• B. $\frac{5}{3}$
• C. $\frac{9}{25}$
• D. $\frac{7}{13}$

Answer 1

The correct option is B

$\frac{5}{3}$

$\frac{x^2+y^2}{x^2-y^2}=\frac{17}{8}$

Dividing numerator and denominator of LHS by $y^2$

$\frac{\frac{x^2}{y^2}+\frac{y^2}{y^2}}{\frac{x^2}{y^2}-\frac{y^2}{y^2}}=\frac{17}{8}$

$\frac{{\left(\frac{x}{y}\right)}^2+1}{{\left(\frac{x}{y}\right)}^2-1}=\frac{17}{8}$

$\Rightarrow 17({\left(\frac{x}{y}\right)}^2-1)=8({\left(\frac{x}{y}\right)}^2+1)$

$\Rightarrow 7{\left(\frac{x}{y}\right)}^2-8{\left(\frac{x}{y}\right)}^2=8+17$

$\Rightarrow 9{\left(\frac{x}{y}\right)}^2=25$

$\Rightarrow {\left(\frac{x}{y}\right)}^2=\frac{25}{9}$

$\Rightarrow \frac{x}{y}=\frac{5}{3}$