Find the equation of the ellipse whose foci is (+-3,0) and which passes through (4,1). Also find the eccentricity and the length of latus rectum of the ellipse.

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Please find below the solution to the asked query:
$$\begin{gathered} \mathrm{Foci} \mathrm{of} \mathrm{ellipse}=\left(3,0\right) \mathrm{and} \left(-3,0\right) \\ \mathrm{As} \mathrm{foci} \mathrm{lie} \mathrm{on} \mathrm{the} \mathrm{X} \mathrm{axis} , \mathrm{hence} \mathrm{equation} \mathrm{of} \mathrm{ellipse} \mathrm{will} \mathrm{be} \\ \Rightarrow \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \left(\mathrm{a}>\mathrm{b}\right) \\ \mathrm{Now} \mathrm{foci} \mathrm{of} \mathrm{above} \mathrm{ellipse} \mathrm{are} \mathrm{S}\left(\mathrm{ae},0\right) \mathrm{and} \mathrm{S}\text{'}\left(-\mathrm{ae},0\right) \\ \Rightarrow \mathrm{ae}=3 \\ \Rightarrow \mathrm{e}=\frac{3}{\mathrm{a}}....\left(1\right) \\ \mathrm{Ellipse} \mathrm{passes} \mathrm{through} \left(4 , 1\right) \\ \mathrm{So} \\ \Rightarrow \frac{4^2}{{\mathrm{a}}^2}+\frac{1^2}{{\mathrm{b}}^2}=1 \\ \Rightarrow \frac{16}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}=1 \\ \mathrm{Because} {\mathrm{b}}^2={\mathrm{a}}^2(1-{\mathrm{e}}^2) \\ \Rightarrow \frac{16}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{a}}^2(1-{\mathrm{e}}^2)}=1 \\ \Rightarrow \frac{1}{{\mathrm{a}}^2}\left(16+\frac{1}{(1-{\mathrm{e}}^2)}\right)=1 \\ \mathrm{Put} \mathrm{e}=\frac{3}{\mathrm{a}} \\ \Rightarrow \frac{1}{{\mathrm{a}}^2}\left(16+\frac{1}{(1-{\displaystyle \frac{3^2}{{\mathrm{a}}^2}})}\right)=1 \\ \Rightarrow 16\left({\mathrm{a}}^2-9\right)+{\mathrm{a}}^2={\mathrm{a}}^2\left({\mathrm{a}}^2-9\right) \\ \Rightarrow {\mathrm{a}}^4-24{\mathrm{a}}^2+144=0 \\ \mathrm{Let} \mathrm{x}={\mathrm{a}}^2 \\ \Rightarrow {\mathrm{x}}^2-24\mathrm{x}+144=0 \\ \Rightarrow \left(\mathrm{x}-16\right)\left(\mathrm{x}-9\right)=0 \\ \Rightarrow \mathrm{x}=16 \mathrm{or} \mathrm{x}=9 \\ \Rightarrow {\mathrm{a}}^2=16 \mathrm{or} {\mathrm{a}}^2=9 \\ \mathrm{So} {\mathrm{b}}^2={\mathrm{a}}^2(1-{\mathrm{e}}^2) \mathrm{and} \mathrm{e}=\frac{3}{\mathrm{a}} \\ \Rightarrow {\mathrm{b}}^2={\mathrm{a}}^2-9 \\ \mathrm{Hence} \\ \Rightarrow {\mathrm{b}}^2=16-9=7 \mathrm{or} {\mathrm{b}}^2=9-9=0 \\ \mathrm{b} \mathrm{can} \mathrm{not} \mathrm{be} \mathrm{zero} \mathrm{so} {\mathrm{a}}^2=16 \mathrm{and} {\mathrm{b}}^2=7 \\ \mathrm{Hence} \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{ellipse} \mathrm{will} \mathrm{be} \\ \Rightarrow \boxed{\frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{7}=1 } \\ \mathrm{Eccentricity} \mathrm{e}=\frac{3}{\mathrm{a}}=\frac{3}{4} \\ \mathrm{and} \mathrm{length} \mathrm{of} \mathrm{latus} \mathrm{rectum} \mathrm{will} \mathrm{be} =\frac{2{\mathrm{b}}^2}{\mathrm{a}}=\frac{2\times 7}{4}=\frac{14}{4}=\frac{7}{2}=3.5 \mathrm{unit} \end{gathered}$$

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