Question
Equation of the ellipse whose axes are the axes of coordinates and which passes through the point $$\left( { - 3,\,1} \right)$$ and has eccentricity $$\sqrt {\frac{2}{5}} $$ is :
A.
$$5{x^2} + 3{y^2} - 48 = 0$$
B.
$$3{x^2} + 5{y^2} - 15 = 0$$
C.
$$5{x^2} + 3{y^2} - 32 = 0$$
D.
$$3{x^2} + 5{y^2} - 32 = 0$$
Answer :
$$3{x^2} + 5{y^2} - 32 = 0$$
Solution :
Let the ellipse be $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$
It passes through $$\left( { - 3,\,1} \right)$$ so $$\frac{9}{{{a^2}}} + \frac{1}{{{b^2}}} = 1.....({\text{i}})$$
Also, $${b^2} = {a^2}\left( {1 - \frac{2}{5}} \right)$$
$$ \Rightarrow 5{b^2} = 3{a^2}.....({\text{ii}})$$
Solving (i) and (ii) we get $${a^2} = \frac{{32}}{3},\,\,\,{b^2} = \frac{{32}}{5}$$
So, the equation of the ellipse is $$3{x^2} + 5{y^2} = 32$$