From a point P , the chord of contact to the ellipse E1 : x^2/3 + y^2/4 = 1 touches the ellipse E2: x^2/9 + y^2/16 = 1 , locus of P satisfy (a,5) then [ a ] = ?? (where [ ] denotes the greatest integer function) .
Answer is 4 .

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \frac{{\mathrm{x}}^2}{3}+\frac{{\mathrm{y}}^2}{4}=1 \\ \mathrm{For} \mathrm{point} \left(\mathrm{h},\mathrm{k}\right) \mathrm{chord} \mathrm{of} \mathrm{contact} \mathrm{to} \mathrm{above} \mathrm{ellipse} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{T}=0 \\ \Rightarrow \frac{\mathrm{xh}}{3}+\frac{\mathrm{yk}}{4}=1 \\ \Rightarrow 4\mathrm{xh}+3\mathrm{yk}=12 \\ \Rightarrow 3\mathrm{yk}=-4\mathrm{xh}+12 \\ \Rightarrow \mathrm{y}=-\frac{4\mathrm{h}}{3\mathrm{k}}\mathrm{x}+\frac{4}{\mathrm{k}} \\ \mathrm{Now} \mathrm{compare} \mathrm{with} \mathrm{y}=\mathrm{mx}+\mathrm{c} \\ \mathrm{m}=-\frac{4\mathrm{h}}{3\mathrm{k}} \mathrm{and} \mathrm{c}=\frac{4}{\mathrm{k}} \\ \mathrm{Now} \mathrm{y}=\mathrm{mx}+\mathrm{c} \mathrm{is} \mathrm{tangent} \mathrm{to} \mathrm{ellipse} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \mathrm{if} {\mathrm{c}}^2={\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2 \\ \mathrm{Hence} \frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{16}=1 \mathrm{will} \mathrm{have} \mathrm{tangent} \mathrm{as} \mathrm{y}=\mathrm{mx}+\mathrm{c} \mathrm{if} \\ {\mathrm{c}}^2=9{\mathrm{m}}^2+16 \\ \Rightarrow \frac{16}{{\mathrm{k}}^2}=9\frac{16{\mathrm{h}}^2}{9{\mathrm{k}}^2}+16 \\ \Rightarrow \frac{1}{{\mathrm{k}}^2}=\frac{{\mathrm{h}}^2}{{\mathrm{k}}^2}+1 \\ \Rightarrow 1={\mathrm{h}}^2+{\mathrm{k}}^2 \\ \mathrm{Hence} \mathrm{locus} \mathrm{is} \\ {\mathrm{x}}^2+{\mathrm{y}}^2=1 \\ \mathrm{Point} \left(\mathrm{a},5\right) \mathrm{cannot} \mathrm{lie} \mathrm{on} \mathrm{this} \mathrm{curve} \mathrm{because} \mathrm{right} \mathrm{hand} \mathrm{can} \mathrm{only} \mathrm{go} \mathrm{upto} 1. \\ Please recheck your question \end{gathered}$$

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