Given, $\frac{x^2}{a}+\frac{x y}{h}+\frac{y^2}{b}=0$
$$
b h x^2+a b x y+a h y^2=0
$$
Two equations are coincident when the ratio of each coefficients is same.
Then, the two required lines are same.
$$
\begin{aligned}
& \text { So, }(y-m x)=0 \\
& \Rightarrow(y-m x)^2=0 \\
& y^2+m^2 x^2-2 m x y=0
\end{aligned}
$$
Compare (i) \& (ii)
$$
b h=m^2 \text { and }-2 m=a b, a h=1
$$
Take $-2 m=a b$ and square both sides
$$
\begin{aligned}
& 4 m^2=a^2 b^2 \\
& 4(b h)=a^2 b^2 \\
& 4 \mathrm{~h}=a^2 b
\end{aligned}
$$
Here, $a=\frac{1}{h}$
Then, $4 h=a \times \frac{1}{h} \times b$
$$
4 h^2=a b
$$
Therefore, option (b) is correct.