The latusrectum of a conic section in the width of the function through the foci. The positive difference between the lengths of the latusrectum of$3\mathrm{y} = {\mathrm{x}}^2 + 4\mathrm{x} – 9$ and ${\mathrm{x}}^2 + 4{\mathrm{y}}^2 – 6\mathrm{x} + 16\mathrm{y} = 24$ is

$$\begin{gathered} \mathrm{given} 3\mathrm{y}={\mathrm{x}}^2+4\mathrm{x}+4-13 \\ {(\mathrm{x}+2)}^2=3\mathrm{y}+13 \\ =3\left(\mathrm{y}+\frac{13}{3}\right) \\ It is parabola so \\ \mathrm{L}.\mathrm{L}.\mathrm{R} =3 \\ {\mathrm{x}}^2+4{\mathrm{y}}^2-6\mathrm{x}+16\mathrm{y}-24=0 \\ ({\mathrm{x}}^2-6\mathrm{x}+9)+4({\mathrm{y}}^2+4\mathrm{y}+4)=24+9+16 \\ {(\mathrm{x}-3)}^2+4{(\mathrm{y}+2)}^2=49 \\ \frac{{(x-3)}^2}{49}+\frac{{(y+2)}^2}{{\displaystyle \frac{49}{4}}}=1 \\ It is an ellipse \\ Now L.L.R=\frac{2b^2}{a} \\ =\frac{2\left(\frac{49}{{\displaystyle 4}}\right)}{7} \\ =\frac{7}{2} \\ Positive difference of L.L.R of given curves=\left|3-\frac{7}{2}\right| =\frac{1}{2} of \end{gathered}$$