Question

The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate axes and passing through the points (4,-1) and (−2,2) is $$$$
A.${\displaystyle \frac{1}{2}}$ $$$$
B. ${\displaystyle \frac{2}{\sqrt{5}}}$$$$$
C. ${\displaystyle \frac{\sqrt{3}}{2}}$$$$$
D. ${\displaystyle \frac{\sqrt{3}}{4}}$$$$$

Answer 1

Hint: Use the given points and the equation of ellipse to find out value of $a^2$and $b^2$ then use them in the formula for eccentricity. $$$$
As given in the question that ellipse is having its center at the origin, so we can write the equation of the ellipse in the form
$${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1...\left(1\right)$$

Complete step-by-step solution:

Here we can see the ellipse cuts $x$ – axis (called major axis) at $(-a,0),(a,0)$ and cuts $y$ –axis at $(0,b),(0,-b)$. The foci of the ellipses are at $(-f,0),(f,0)$. The eccentricity is defined as the ratio of distance between two foci and the size of the major axis. In this case, $e={\displaystyle \frac{2f}{2a}}={\displaystyle \frac{f}{a}}$. $$$$
Where we know that $a$ is the length of semi-major axis and $b$ is the length of semi-minor axis. The ellipse passes through the points (4,-1) and (−2,2) . So these must satisfy equation (1) . Putting each point in equation(1) we get a pair of equations
$$\begin{array}{rl}& {\displaystyle \frac{16}{a^2}}+{\displaystyle \frac{1}{b^2}}=1...\left(2\right)\\ & {\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{1}{b^2}}={\displaystyle \frac{1}{4}}...\left(3\right)\end{array}$$
Let us substitute $X={\displaystyle \frac{1}{a^2}},Y={\displaystyle \frac{1}{b^2}}$ in equation (3) and (2) to convert them into a pair of linear equations
$$\begin{array}{rl}& 16X+Y=1\dots \dots \dots \dots \dots \dots \dots \dots .\left(4\right)\\ & X+Y={\displaystyle \frac{1}{4}}\dots \dots \dots \dots \dots \dots \dots .\left(5\right)\end{array}$$
We solve linear system of equations by subtracting equation (5) from (4) to get,
$$\begin{array}{rl}& 15X={\displaystyle \frac{3}{4}}\\ & \Rightarrow X={\displaystyle \frac{1}{20}}\\ & \Rightarrow a^2=20\end{array}$$
Putting the value of $X$ in equation (5) we get
$$\begin{array}{rl}& Y={\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{20}}\\ & \Rightarrow Y={\displaystyle \frac{1}{5}}\\ & \Rightarrow b^2=5\end{array}$$
The eccentricity of an ellipse defines its shape and size. Ellipse has more than one eccentricity. If an eccentricity increases then the shape is more spherical and if it decreases then the shape is less spherical. The eccentricity with respect to semi-major axis is given by $e=\sqrt{1-{\displaystyle \frac{b^2}{a^2}}}$. The value of eccentricity varies between strictly 0 and 1. Let us put the values we obtained from solution of pair of equations to find eccentricity,
$$e=\sqrt{1-{\displaystyle \frac{b^2}{a^2}}}=\sqrt{1-{\displaystyle \frac{5}{20}}}={\displaystyle \frac{\sqrt{3}}{2}}$$
 We have rejected the negative value as eccentricity lies between strictly 0 and 1 for ellipse. So the correct choice is C.

Note: The question tests your knowledge of the eccentricity and its formula in two dimensions. The problem can also be reframed to ask to find other eccentricity or the range of eccentricity with respect to a locus.