Answer:
End point of Major Axis =(-11,4) and (9,4), as ends points should have same but opposite in sign x coordinate.So, either it should be (-11,4) and (11,4) or (-9,4) and (9,4).
Length of Major axis ,as y coordinates are same.
= |9-(-9)| or |11-(-11)|
=18 or 22
Let equation of ellipse be
[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\\\2a=18\\\\a=9\\\\\text{Ellipse passes through} (7,7)\\\\\frac{7^2}{9^2}+\frac{7^2}{b^2}=1\\\\\frac{49}{b^2}=1-\frac{49}{81}\\\\\frac{49}{b^2}=\frac{32}{81}\\\\b^2=\frac{49*81}{32}\\\\b=\pm \frac{63}{\sqrt{32}}\\\\2a=22\\\\a=11\\\\\text{Ellipse passes through} (7,7)\\\\\frac{7^2}{11^2}+\frac{7^2}{b^2}=1\\\\\frac{49}{b^2}=1-\frac{49}{121}\\\\\frac{49}{b^2}=\frac{72}{81}\\\\b^2=\frac{49*81}{72}\\\\b=\pm \frac{63}{\sqrt{72}}[/tex]
Equation of ellipse will be
[tex]\rightarrow \frac{x^2}{9^2}+\frac{y^2}{(\pm \frac{63}{\sqrt{32}})^2}=1\\\\\rightarrow \frac{x^2}{81}+\frac{32y^2}{3969}=1[/tex]
Area of the ellipse in first Quadrant , if coordinates of end points of major axis are (-9,4) and (9,4).
[tex]=\int\limits^9_0 {y} \, dx\\\\=\int\limits^9_0 {\sqrt{\frac{3969}{32}\times (1-\frac{x^2}{81})} \, dx\\\\=\sqrt{\frac{3969}{32*81}} \times\int\limits^9_0 {\sqrt{(9^2-x^2}\\\\=\sqrt{\frac{3969}{2592}} \times (\frac{x*\sqrt{9^2-x^2}}{2}+\frac{81}{2}\sin^{-1}{\frac{x}{9})\left \{ {{x=9} \atop {x=0}} \right[/tex]
[tex]=\sqrt{\frac{3969}{2592}} \times \frac{81}{2} \times \frac{\pi}{2}\\\\=\sqrt{\frac{3969}{2592}} \times \frac{81\pi}{4}[/tex]
So, Area of Ellipse
=4 × Area of ellipse in first Quadrant
[tex]=\sqrt{\frac{3969}{2592}}\times 81\pi[/tex]
If, coordinates of end points of major axis is (-11,4) and (11,4), evaluate by same procedure.
