Question

Find the equation to the ellipse (referred to its axes as the axes of x and y respectively) which passes through the point (−3, 1) and has eccentricity \[\sqrt{\frac{2}{5}}\]

 

Answer 1

\[\text{ Let the equation of the ellipse be } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ...(1)\]
\[\text{ It passes through the point} \left( -3,1 \right).\]
\[\therefore\frac{9}{a^2}+\frac{1}{b^2}=1 ... (2)\]
\[\text{ Also, } e = \sqrt{\frac{2}{5}}\]
\[\text{ Now, } b^2 = a^2 \left( 1 - e^2 \right)\]
\[ \Rightarrow b^2 = a^2 \left[ 1 - \frac{2}{5} \right]\]
\[ \Rightarrow b^2 = \frac{3 a^2}{5}\]
\[\text{ Substituting the value of } b^2 \text{ in eq. (2), we get } :\]
\[\frac{9}{a^2}+\frac{5}{3 a^2}=1\]
\[ \Rightarrow \frac{27 + 5}{3 a^2} = 1\]
\[ \Rightarrow a^2 = \frac{32}{3}\]
\[ \Rightarrow b^2 = \frac{3 \times \frac{32}{3}}{5} \text{ or } \frac{32}{5}\]
\[\text{ Substituting the values of a and b in eq. (1), we get } :\]
\[\frac{3 x^2}{32} + \frac{5 y^2}{32} = 1\]
\[ \Rightarrow 3 x^2 + 5 y^2 = 32\]
\[\text{ This is the required equation of the ellipse.} \]