Question

Q.34. Find the equation of the ellipse whose center at origin, major axis on the X axis and passes through the point (4, 3) and (6, 2).

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{ellipse}\mathrm{having}\mathrm{the}\mathrm{centre}\mathrm{at}\mathrm{the}\mathrm{origin}\left(0,0\right)\mathrm{and}\mathrm{having}\mathrm{major}\mathrm{axis}\mathrm{on}\mathrm{x}-\mathrm{axis}\mathrm{is} \\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1.....\left(1\right) \\ \mathrm{Here},\mathrm{a}=\mathrm{length}\mathrm{of}\mathrm{semi}-\mathrm{major}\mathrm{axis};\mathrm{b}=\mathrm{length}\mathrm{of}\mathrm{semi}-\mathrm{minor}\mathrm{axis} \\ \mathrm{Now},\left(1\right)\mathrm{passes}\mathrm{through}\left(4,3\right),\mathrm{so}\mathrm{it}\mathrm{must}\mathrm{satisfy}\mathrm{it}. \\ \frac{16}{{\mathrm{a}}^2}+\frac{9}{{\mathrm{b}}^2}=1......\left(2\right) \\ \mathrm{Now},\left(1\right)\mathrm{passes}\mathrm{through}\left(6,2\right),\mathrm{so}\mathrm{it}\mathrm{must}\mathrm{satisfy}\mathrm{it}. \\ \frac{36}{{\mathrm{a}}^2}+\frac{4}{{\mathrm{b}}^2}=1......\left(3\right) \\ \mathrm{Let}\frac{1}{{\mathrm{a}}^2}=\mathrm{u}\mathrm{and}\frac{1}{{\mathrm{b}}^2}=\mathrm{v} \\ \mathrm{Now},\left(2\right)\mathrm{and}\left(3\right)\mathrm{becomes}, \\ 16\mathrm{u}+9\mathrm{v}=1.....\left(4\right) \\ 36\mathrm{u}+4\mathrm{v}=1......\left(5\right) \\ \mathrm{Multiply}\left(4\right)\mathrm{by}4\mathrm{and}\left(5\right)\mathrm{by}9,\mathrm{we}\mathrm{get} \\ 64\mathrm{u}+36\mathrm{v}=4......\left(7\right) \\ 324\mathrm{u}+36\mathrm{v}=9....\left(8\right) \\ \mathrm{Subtracting}\left(7\right)\mathrm{from}\left(8\right),\mathrm{we}\mathrm{get} \\ 260\mathrm{u}=5 \\ \Rightarrow \mathrm{u}=\frac{5}{260}=\frac{1}{52} \\ \mathrm{Put}\mathrm{u}=\frac{1}{52}\mathrm{in}\left(7\right),\mathrm{we}\mathrm{get} \\ 64\times \frac{1}{52}+36\mathrm{v}=4 \\ \Rightarrow \frac{16}{13}+36\mathrm{v}=4 \\ \Rightarrow 36\mathrm{v}=4-\frac{16}{13} \\ \Rightarrow 36\mathrm{v}=\frac{36}{13} \\ \Rightarrow \mathrm{v}=\frac{1}{13} \\ \mathrm{So},{\mathrm{a}}^2=52\mathrm{and}{\mathrm{b}}^2=13 \\ \mathrm{The}\mathrm{required}\mathrm{equation}\mathrm{of}\mathrm{ellipse}\mathrm{is}, \\ \boxed{\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{52}}\mathbf{}\mathbf{+}\mathbf{}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{13}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}} \end{gathered}$$