An ellipse has foci at (9, 20) and (49, 55) in the x-y plane and is tangent to x-axis. Let the length of its major axis is L, then $\left[\frac{\mathrm{L}}{11}\right]$= (where [.] = G.I.F)

$\begin{array}{l}\mathrm{Given} \mathrm{foci} \mathrm{are} {\mathrm{F}}_1(9,20),F_2(49,55)\end{array}$

Now image of $F_1 w.r.to \tan gent x-axis is F_1\text{'}=(9,-20)$

Now the points $F_2,P,F_1\text{'} are collinear$

$$\begin{gathered} Now 2a=F_{1}P+F_{2}P=F{\text{'}}_{1}P+F_{2}P=F_1\text{'}{F}_2 \\ \Rightarrow 2a=\sqrt{{(49-9)}^2+{(55+20)}^2}=\sqrt{7225}=85 \\ length of major axis\left(L\right)=85 \\ Now \left[\frac{L}{11}\right]=\left[\frac{85}{11}\right]=7 \end{gathered}$$