Equation of the ellipse whose axes of coordinates and which passes through the point (−3,1) and has eccentricity \[\sqrt {{2 \over 5}} \] is :
Solutions · 1
Official solution
Let the ellipse be \[{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1\]
It press through (−3,1) so \[{9 \over {{a^2}}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( i \right)\]
Also, \[{b^2} = {a^2}\left( {1 - 2/5} \right)\]
\[ \Rightarrow 5{b^2} = 3{a^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)\]
Solving (i) and (ii) we get \[{a^2} = {{32} \over 3},{b^2} = {{32} \over 5}\]
So, the equation of the ellipse is 3x2+5y2=32