If l, m are variable real number such that $5{\mathrm{l}}^2+6{\mathrm{m}}^2-4\mathrm{lm}+3\mathrm{l}=0$, then variable line lx + my = 1 always touches a fixed parabola, whose axis is parallel to x-axis

 Let EQ of parabola is $(\mathrm{y}-\mathrm{k}{)}^2=4\mathrm{a}(\mathrm{x}-\mathrm{h})$
Equation of tangent is $\mathrm{y}-\mathrm{k}={\mathrm{m}}_1(\mathrm{x}-\mathrm{h})+\mathrm{a}/{\mathrm{m}}_1$
Comparing with given tgt $\mathrm{y}=-\frac{\mathrm{l}}{\mathrm{m}}\mathrm{x}+\frac{1}{\mathrm{n}}$
And eliminate m₁ we get ${\mathrm{hl}}^2-{\mathrm{am}}^2+\mathrm{klm}-1$
Now comparing with
$\begin{array}{l}\left(\frac{-5}{3}\right){\mathrm{l}}^2-2{\mathrm{m}}^2+\frac{4}{3}\mathrm{lm}-\mathrm{l}=0\\ \Rightarrow \mathrm{h}=\frac{-5}{3},\mathrm{a}=2,\mathrm{k}=\frac{4}{3}\end{array}$
Vertex $\mathrm{v}(\mathrm{h},\mathrm{k})=\left(\frac{-5}{3},\frac{4}{3}\right)$
Focus = $=\mathrm{s}\left(\frac{1}{3},\frac{4}{3}\right)$
Equation of diretrix is 3x + 11 = 0.