If \(\mathrm{m}\) is the slope of one of the lines represented by \(a x^{2}+2 h x y+b^{2}=0\), then \((h+b m)^{2}\) is equal to

Given that, \(a x^{2}+2 h x y+b y^{2}=0 \quad \text{...(i)}\)
Which is homogeneous equation representing pair of straight line each of which passing through the origin. Given one slope of line \(=\mathrm{m}\).
Let another slope of line \(=m_{1}\)
Then, the lines are \(y=m x\) and \(y=m_{1} x\)
Now, \(\quad(\mathrm{mx}-\mathrm{y})\left(\mathrm{m}_{1} \mathrm{x}-\mathrm{y}\right)\)
\(\Rightarrow \quad \mathrm{mm}_{1} \mathrm{x}^{2}-\mathrm{m}_{1} \mathrm{xy}-\mathrm{mxy}+\mathrm{y}^{2}\) \(\Rightarrow \quad \quad \mathrm{mm}_{1} \cdot \mathrm{x}^{2}-\left(\mathrm{m}+\mathrm{m}_{1}\right) \mathrm{y} \cdot \mathrm{x}+\mathrm{y}^{2} \quad \text{...(ii)}\)
On comparing Eqs. (i) and (ii),
\[
\begin{aligned}
\mathrm{m}+\mathrm{m}_{1} &=-\frac{2 \mathrm{~h}}{\mathrm{~b}} \quad \text{...(iii)} \\
\mathrm{~mm}_{1} &=\frac{\mathrm{a}}{\mathrm{b}} \quad \text{...(iv)}
\end{aligned}
\]
From Eqs. (iii) and (iv),
\[
\mathrm{m}_{1}=\left(-\frac{2 \mathrm{~h}}{\mathrm{~b}}-\mathrm{m}\right)
\]
\(\begin{array}{lc}\Rightarrow & \mathrm{m}\left(\frac{-2 \mathrm{~h}}{\mathrm{~b}}-\mathrm{m}\right)=\frac{\mathrm{a}}{\mathrm{b}} \\ \Rightarrow & -\frac{\mathrm{m}}{\mathrm{b}}(2 \mathrm{~h}+\mathrm{mb})=\frac{\mathrm{a}}{\mathrm{b}} \\ \Rightarrow & -2 \mathrm{mh}-\mathrm{m}^{2} \mathrm{~b}=\mathrm{a} \\ \Rightarrow & -2 \mathrm{mhb}-\mathrm{m}^{2} \mathrm{~b}^{2}=\mathrm{ab} \\ \Rightarrow & \mathrm{h}^{2}+2 \mathrm{mhb}+\mathrm{m}^{2} \mathrm{~b}^{2}=-\mathrm{ab}+\mathrm{h}^{2} \\ \Rightarrow & (\mathrm{h}+\mathrm{mb})^{2}=\mathrm{h}^{2}-\mathrm{ab}\end{array}\)