Question

Equation $x = {\text{ }}a{\text{ }}\cos \theta$ and $y = b\sin \theta$ represents a conic section whose eccentricity $e$is given by

$${\text{A}}{\text{. }}{e^2} = \dfrac{{{a^2} + {b^2}}}{{{a^2}}} \\\ {\text{B}}{\text{. }}{e^2} = \dfrac{{{a^2} + {b^2}}}{{{b^2}}} \\\ {\text{C}}{\text{. }}{e^2} = \dfrac{{{a^2} - {b^2}}}{{{a^2}}}{\text{ }} \\\$$

${\text{D}}{\text{.}}$None of these

Answer 1

In order to solve this problem we need to use the Trigonometric identity ${\cos ^2}\theta + {\sin ^2}\theta = 1$. We also need to know the general formula of the equation of the ellipse and the formula to find the eccentricity of the ellipse.

Complete step-by-step answer:
As in the question it is given that $x = {\text{ }}a{\text{ }}\cos \theta$ and $y = b\sin \theta$. So first we will rearrange the equation as below-
$\cos \theta = \dfrac{x}{a}$ and $\sin \theta = \dfrac{y}{b}$ now if we cleverly look at this two obtained expression we find that if we square both the sides and add both the expression together like -
$\Rightarrow {\cos ^2}\theta = {\left( {\dfrac{x}{a}} \right)^2}$and ${\sin ^2}\theta = {\left( {\dfrac{y}{a}} \right)^2}$
$\Rightarrow {\cos ^2}\theta + {\sin ^2}\theta = {\left( {\dfrac{x}{a}} \right)^2} + {\left( {\dfrac{y}{b}} \right)^2}$ and we know that ${\cos ^2}\theta + {\sin ^2}\theta = 1$
So, ${\left( {\dfrac{x}{a}} \right)^2} + {\left( {\dfrac{y}{b}} \right)^2} = 1$
Alternate solution,
We know that
${\cos ^2}\theta + {\sin ^2}\theta = 1{\text{ }}........\left( 1 \right)$
So Put the value of $\cos \theta = \dfrac{x}{a}$ and $\sin \theta = \dfrac{y}{b}$ as obtained in equation(1).
Therefore, we get
$\Rightarrow {\cos ^2}\theta + {\sin ^2}\theta = 1 \\\ \Rightarrow \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1 \\\$
$\Rightarrow {\left( {\dfrac{x}{a}} \right)^2} + {\left( {\dfrac{y}{b}} \right)^2} = 1$
By seeing the equation, it represents an equation of ellipse centered at origin and with axes lying along the coordinate axes and we know that eccentricity of ellipse is given by
$e = \sqrt {1 - \dfrac{{{b^2}}}{{{a^2}}}}$ where $e$ represents eccentricity of an ellipse.
Eccentricity is the ratio of the distance from the centre to the foci and the distance from the centre to the vertices.
On further rearranging the above expression we get
$\Rightarrow {e^2} = \dfrac{{{a^2} - {b^2}}}{{{a^2}}}$

So, the correct answer is “Option C”.

Note: Whenever this type of question appears always note down the given things first. After this using trigonometric identity and putting the given values we will get the desired expression. Remember $e$ represents the eccentricity of an ellipse. Eccentricity is the ratio of the distance from the centre to the foci and the distance from the centre to the vertices. As Obtained in solution eccentricity of ellipse is given as ${e^2} = \dfrac{{{a^2} - {b^2}}}{{{a^2}}}$. Study more about major axis, minor axis, foci, vertex and latus rectum of ellipse.