The given equation of ellipse is,
[tex]\frac{(x-2)^2}{16}+\frac{y^2}{4}=1\text{ ---(1)}[/tex]
The above equation can be rewritten as,
[tex]\frac{(x-2)^2}{4^2}+\frac{y^2}{2^2}=1\text{ ----(2)}[/tex]
The above equation is similar to the standard form of the ellipse with center (h, k) and major axis parallel to x axis given by,
[tex]\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1\text{ ----(3)}[/tex]
where a>b.
Comparing equations (2) and (3), h=2, k=0, a=4 and b= 2.
Hence, the center of the ellipse is (h, k)=(2, 0).
The coordinates of the vertices are given by,
[tex]\begin{gathered} (h+a,\text{ k)=(2+}4,\text{ }0)=(6,\text{ 0)} \\ (h-a,\text{ k)=(2-}4,\text{ }0)=(-2,\text{ 0)} \end{gathered}[/tex]
Hence, the coordinates of the vertices are (6, 0) and (-2,0).
The coordinates of the co-vertices are given by,
[tex]\begin{gathered} (h,\text{ k+}b)=(2,\text{ }0+2)=(2,\text{ 2)} \\ (h,\text{ k-}b)=(2,\text{ }0-2)=(2,\text{ -2)} \end{gathered}[/tex]
Hence, the coordinates of the co-vertices are (2, 2) and (2, -2).
The coordinates of the foci are (h±c, k).
[tex]\begin{gathered} c^2=a^2-b^2 \\ c^2=4^2-2^2 \\ c^2=16-4 \\ c^2=12 \\ c=2\sqrt[]{3} \end{gathered}[/tex]
Using the value of c, the coordinates of the foci are,
[tex]\begin{gathered} \mleft(h+c,k\mright)=(2+2\sqrt[]{3},\text{ 0)} \\ (h-c,k)=(2-2\sqrt[]{3},\text{ 0)} \end{gathered}[/tex]
Therefore, the coordinates of the foci are,
[tex](2+2\sqrt[]{3},\text{ 0) and }(2-2\sqrt[]{3},\text{ 0)}[/tex]
The endpoints of the latus rectum is,
[tex]\begin{gathered} (h+c,\text{ k}+\frac{b^2}{a})=(2+2\sqrt[]{3},\text{ 0+}\frac{2^2}{4^{}}) \\ =(2+2\sqrt[]{3},\text{ 1)}^{} \\ (h-c,\text{ k}+\frac{b^2}{a})=2-2\sqrt[]{3},\text{ 0+}\frac{2^2}{4^{}}) \\ =(2-2\sqrt[]{3},\text{ 1}^{}) \\ (h+c,\text{ k-}\frac{b^2}{a})=(2+2\sqrt[]{3},\text{ 0-}\frac{2^2}{4^{}}) \\ =(2+2\sqrt[]{3},\text{ -1}^{}) \\ (h-c,\text{ k-}\frac{b^2}{a})=(2-2\sqrt[]{3},\text{ 0-}\frac{2^2}{4^{}}) \\ =(2-2\sqrt[]{3},\text{ -1}^{}) \end{gathered}[/tex]
Therefore, the coordinates of the end points of the latus recta is,
[tex](2+2\sqrt[]{3},\text{ 1)},\text{ }(2-2\sqrt[]{3},\text{ 1}^{}),\text{ }(2+2\sqrt[]{3},\text{ -1}^{})\text{ and }(2-2\sqrt[]{3},\text{ -1}^{})[/tex]
Now, the equations of the directrices is,
[tex]\begin{gathered} x=h\pm\frac{a}{e} \\ x=\pm\frac{a}{\sqrt[]{1-\frac{b^2}{a^2}}} \\ x=2\pm\frac{4}{\sqrt[]{1-\frac{2^2}{4^2}}} \\ x=2\pm\frac{4}{\sqrt[]{1-\frac{1^{}}{4^{}}}} \\ x=2\pm\frac{4}{\sqrt[]{\frac{3}{4}^{}}} \\ x=2\pm4\sqrt[]{\frac{4}{3}} \end{gathered}[/tex]
Here, e is the eccentricity of the ellipse.
Therefore, the directrices of the ellipse is
[tex]x=2\pm4\sqrt[]{\frac{4}{3}}[/tex]
Now, the graph of the ellipse is given by,
