Question

If \[{b^2} = {a^2}\left( {1 - {e^2}} \right)\], then \[e\] is

Answer 1

Hint: To find the value of \[e\], perform operations such that only \[e\] remain on one side of the equation and rest terms should be there on the opposite side.

Complete step-by-step answer:
Given, \[{b^2} = {a^2}\left( {1 - {e^2}} \right)\]
Divide both sides of the equation by \[{a^2}\].
\[\dfrac{{{b^2}}}{{{a^2}}} = \dfrac{{{a^2}\left( {1 - {e^2}} \right)}}{{{a^2}}}\]
Factor out the common factor.
\[\dfrac{{{b^2}}}{{{a^2}}} = \left( {1 - {e^2}} \right)\]
Subtract 1 from both sides of the equation.
\[\dfrac{{{b^2}}}{{{a^2}}} - 1 = \left( {1 - {e^2}} \right) - 1\]
Simply the obtained equation.
\[\dfrac{{{b^2}}}{{{a^2}}} - 1 = - {e^2}\]
Multiply both sides by \[ - 1\]and simplify.
\[
   - 1\left( {\dfrac{{{b^2}}}{{{a^2}}} - 1} \right) = - 1\left( { - {e^2}} \right) \\
  {e^2} = 1 - \dfrac{{{b^2}}}{{{a^2}}} \\
 \]
Take the square root of both sides of the equation to obtain the value of \[e\].
\[e = \pm \sqrt {1 - \dfrac{{{b^2}}}{{{a^2}}}} \]

Note: Whenever square root is performed, there are two results, one positive and other negative, therefore it is compulsory to put \[ \pm \]before the result. Common mistake made by students in these type of questions is when values are shifted from left to right or right to left of the equal sign, therefore whenever moving a value from right to left or left to right the sign always changes from \[ + \to - \] and \[ - \to + \]