Step 1: Use the standard result:
Write the equation as $Ax^2+2Hxy+By^2=0$ with $A=\frac1a$, $H=\frac1h$ and $B=\frac1b$. For a pair of lines with slopes in ratio $1:2$ the known condition is $8H^2=9AB$.
Step 2: Recall why:
For slopes $m_1,m_2$ we have $m_1+m_2=-\frac{2H}{B}$ and $m_1m_2=\frac{A}{B}$. With $m_2=2m_1$, then $\frac{(m_1+m_2)^2}{m_1m_2}=\frac{9m_1^2}{2m_1^2}=\frac92$. So $\frac{4H^2/B^2}{A/B}=\frac{4H^2}{AB}=\frac92$, giving $8H^2=9AB$.
Step 3: Substitute:
$8\cdot\frac{1}{h^2}=9\cdot\frac{1}{a}\cdot\frac{1}{b}$, so $\frac{8}{h^2}=\frac{9}{ab}$.
Step 4: Read off the ratio:
Cross-multiplying gives $9h^2=8ab$, so $\frac{h^2}{ab}=\frac{8}{9}$.
Final Answer:
We get $h^2:ab=8:9$ (option C).
\[ \boxed{8:9} \]