Question

What is the equation of the ellipse whose axes are the axes of coordinates and which passes through the point $( - 3,1)$ and has eccentricity $\sqrt {\dfrac{2}{5}} $ is:
A. $3{x^2} + 5{y^2} - 32 = 0$
B. $5{x^2} + 3{y^2} - 48 = 0$
C. $3{x^2} + 5{y^2} - 15 = 0$
D. $5{x^2} + 3{y^2} - 32 = 0$

Answer 1

Hint:We know the standard equation of ellipse, that is, \[{\left( {\dfrac{x}{a}} \right)^2} + {\left( {\dfrac{y}{b}} \right)^2} = 1\] and eccentricity of ellipse is $e = \sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} $ . We are also given with one point through which ellipse passes which we can utilize in the standard equation of ellipse. So we are available with two equations, and two unknowns, that are, a and b. We can easily find the answer.

Complete step-by-step answer:
Ellipse: It is a conic section formed by the intersection of a right circular cone by a plane that cuts the axis and the surface of the cone.
We know standard equation of ellipse, that is \[{\left( {\dfrac{x}{a}} \right)^2} + {\left( {\dfrac{y}{b}} \right)^2} = 1\].
Substituting the given point in standard equation of ellipse we get,
\[{\left( {\dfrac{{ - 3}}{a}} \right)^2} + {\left( {\dfrac{1}{b}} \right)^2} = 1\] ---------(1)
The eccentricity (e) of an ellipse is the ratio of the distance from the center to the foci (c) and the distance from the center to the vertices (a).
We know eccentricity of an ellipse is given by,
\[e = \sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} \]
Squaring on both sides,
${e^2} = \dfrac{{{a^2} - {b^2}}}{{{a^2}}}$
On putting value of eccentricity, we get,
$
  \dfrac{2}{5} = \dfrac{{{a^2} - {b^2}}}{{{a^2}}} \\
    \\
 $
On further solving,
$2{a^2} = 5{a^2} - 5{b^2}$
So, we get
${a^2} = \dfrac{5}{3}{b^2}$
Putting this value of ${a^2}$ in equation (1), we get,
\[\left( {\dfrac{{27}}{{5{b^2}}}} \right) + \left( {\dfrac{1}{{{b^2}}}} \right) = 1\]
On further solving, we get,
\[5{b^2} = 32\]
So we get the result,
\[{b^2} = \dfrac{{32}}{5}\]
Using this,
\[{a^2} = \dfrac{{5{b^2}}}{3} \Rightarrow {a^2} = \dfrac{{32}}{3}\]
Putting value of ${a^2}$ and ${b^2}$ in standard equation of ellipse, we get,
$\dfrac{{3{x^2}}}{{32}} + \dfrac{{5{y^2}}}{{32}} = 1 \Rightarrow 3{x^2} + 5{y^2} - 32 = 0$

So, the correct answer is “Option A”.

Note:We noticed that, value of a is greater than b as we know the value of a is always greater than b in an ellipse. We have used the given point, which was passing through the ellipse, in the standard equation of the ellipse to get one equation and the property of eccentricity and it’s given value in the question to get the second equation and then we solved both equations to get our answer.