If the slopes of the lines given by the equation \(a x^{2}+2 h x y+b y^{2}=0 \quad\) are in the
ratio \(5: 3\), then the ratio \(h^{2}: a b=\)

(D)
Let \(y=m_{1} x\) and \(y=m_{2} x\) be the lines represented by the equation. \(a x^{2}+2 h x y+b y^{2}=0\)
Then, \(m_{1}+m_{2}=\frac{-2 h}{b}\) and \(m_{1} m_{2}=\frac{a}{b}\)
We have, \(\frac{m_{1}}{m_{2}}=\frac{5}{3} \Rightarrow m_{1} \Rightarrow \frac{5 m_{2}}{3}\)
\(\therefore \frac{5 \mathrm{~m}_{2}}{3}+\mathrm{m}_{2}=\frac{-2 \mathrm{~h}}{\mathrm{~b}}\) and \(\left(\frac{5 \mathrm{~m}_{2}}{3}\right) \mathrm{m}_{2}=\frac{\mathrm{a}}{\mathrm{b}}\)
\(\therefore \frac{8 m_{2}}{3}=\frac{-2 h}{b} \Rightarrow m_{2}=\frac{-3 h}{4 b}\) and \(m_{2}^{2}=\frac{3 a}{5 b}\)
\(\left(\frac{-3 h}{4 b}\right)^{2}=\frac{3 a}{5 b} \Rightarrow \frac{9 h^{2}}{16 b^{2}}=\frac{3 a}{5 b}\)
\(\therefore \frac{\mathrm{h}^{2}}{\mathrm{ab}}=\frac{16}{15}\)