Question

Find the eccentricity of the conic \[\left[ {\dfrac{{{{(x + 2)}^2}}}{7}} \right] + {\left( {y - 1} \right)^2} = 14\] .
A. \[\sqrt {\dfrac{7}{8}} \]
B. \[\sqrt {\dfrac{6}{{17}}} \]
C. \[\sqrt {\dfrac{3}{2}} \]
D. \[\sqrt {\dfrac{6}{{11}}} \]
E. \[\sqrt {\dfrac{6}{7}} \]

Answer 1

Hint: The equation provided in the question can be transformed into an equation of ellipse from which we will get \[a\] and \[b\] . Then the value of \[a\]and \[b\] we shall put it in the formula of eccentricity.

Formula Used:
The formula of eccentricity is \[\sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} \] , where a and b are the constants of the general equation of the ellipse.

Complete step by step solution:
We have to write the equation given in the question in the form of the general equation of ellipse which is \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\] .
Simplifying the equation into the general equation we will get –
\[\left[ {\dfrac{{{{(x + 2)}^2}}}{7}} \right] + {\left( {y - 1} \right)^2} = 14\]
\[\dfrac{{{{\left( {x + 2} \right)}^2}}}{{{{\left( {\sqrt {98} } \right)}^2}}} + \dfrac{{{{\left( {y - 1} \right)}^2}}}{{{{\left( {\sqrt {14} } \right)}^2}}} = 1\]
\[\dfrac{{{{\left( {x + 2} \right)}^2}}}{{{{\left( {7\sqrt 2 } \right)}^2}}} + \dfrac{{{{\left( {y - 1} \right)}^2}}}{{{{\left( {\sqrt {14} } \right)}^2}}} = 1\] ------ (i)
This is the equation of ellipse from which we will get the values of \[a\] and \[b\] which we will be substituting in the formula of eccentricity.
Now from the equation (i) we will get the values of \[a\] and \[b\] as –
\[a = 7\sqrt 2 \] and \[b = \sqrt {14} \] .
Putting the values of \[a\] and \[b\] in the formula we will get –
Eccentricity =\[\sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} \]
\[ = \sqrt {\dfrac{{{{\left( {7\sqrt 2 } \right)}^2} - {{\left( {\sqrt {14} } \right)}^2}}}{{{{\left( {7\sqrt 2 } \right)}^2}}}} \]
\[ = \sqrt {\dfrac{{98 - 14}}{{98}}} \]
\[ = \sqrt {\dfrac{{84}}{{98}}} \]
Dividing by 2 from numerator and denominator, we get
\[ = \sqrt {\dfrac{{42}}{{49}}} \]
Dividing by 2 from numerator and denominator, we get
\[ = \sqrt {\dfrac{6}{7}} \]
Hence the answer is (E) which is \[ = \sqrt {\dfrac{6}{7}} \]

Note: Students often confuse with formula of eccentricity of a hyperbola and ellipse. The formula of eccentricity of ellipse is \[\sqrt {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} \] .