Question

Qno.15

$15.Findtheequationofanellipsewhoseaxesliealongthecoordinateaxes,whichpassesthroughthepoint(-3,1)andhaseccentricityequalto\sqrt{2/5}.$

Answer 1

Dear student
$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{ellipse}\mathrm{be}\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1...\left(1\right) \\ \mathrm{It}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}(-3,1) \\ \therefore \frac{(-3{)}^2}{{\mathrm{a}}^2}+\frac{(1{)}^2}{{\mathrm{b}}^2}=1 \\ \Rightarrow \frac{9}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}=1...\left(2\right) \\ \mathrm{and},\mathrm{e}=\sqrt{\frac{2}{5}} \\ \mathrm{Now}, \\ {\mathrm{b}}^2={\mathrm{a}}^2(1-{\mathrm{e}}^2) \\ \Rightarrow {\mathrm{b}}^2={\mathrm{a}}^2\left(1-{\left(\sqrt{\frac{2}{5}}\right)}^2\right) \\ \Rightarrow {\mathrm{b}}^2={\mathrm{a}}^2\left(1-\frac{2}{5}\right) \\ \Rightarrow {\mathrm{b}}^2={\mathrm{a}}^2\left(\frac{5-2}{5}\right) \\ \Rightarrow {\mathrm{b}}^2={\mathrm{a}}^2\left(\frac{3}{5}\right) \\ \Rightarrow {\mathrm{b}}^2=\frac{3{\mathrm{a}}^2}{5}...\left(3\right) \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{b}}^2\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ \frac{9}{{\mathrm{a}}^2}+\frac{5}{3{\mathrm{a}}^2}=1 \\ \Rightarrow \frac{27+5}{3{\mathrm{a}}^2}=1 \\ \Rightarrow {\mathrm{a}}^2=\frac{32}{3} \\ \mathrm{Put}\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{a}}^2\mathrm{in}\left(3\right),\mathrm{we}\mathrm{get} \\ {\mathrm{b}}^2=\frac{3\times \frac{32}{3}}{5} \\ \Rightarrow {\mathrm{b}}^2=\frac{32}{5} \\ \mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \frac{3{\mathrm{x}}^2}{32}+\frac{5{\mathrm{y}}^2}{32}=1 \\ \Rightarrow \boxed{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{5}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{32}}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{equation}\mathrm{of}\mathrm{ellipse} \end{gathered}$$
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