Question

If the eccentricity \[{e_1}\] is of the ellipse \[\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{{25}} = 1\] and \[{e_2}\] is the eccentricity of the hyperbola passing through the foci of the ellipse and \[{e_1} \times {e_2} = 1\], then the equation of the hyperbola. Is
A). \[\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{{16}} = 1\]
B). \[\dfrac{{{x^2}}}{{16}} - \dfrac{{{y^2}}}{9} = - 1\]
C). \[\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{{25}} = 1\]
D). None of these

Answer 1

Hint: To solve this question first we find the eccentricity of the ellipse. Then we find the eccentricity of the hyperbola through the given relation of multiplication of the eccentricity. Then we use the relation of passing the hyperbola through the foci of the ellipse. And obtained the equation of the hyperbola.

Complete step-by-step solution:
Equation of the ellipse \[\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{{25}} = 1\]
\[{e_1}\] is the eccentricity of the ellipse.
Formula of the eccentricity of the ellipse is \[e = \sqrt {1 - \dfrac{{{a^2}}}{{{b^2}}}} \]
On putting the values from the equation of the ellipse.
\[{e_1} = \sqrt {1 - \dfrac{{16}}{{25}}} \]
On further solving
\[{e_1} = \sqrt {\dfrac{{25 - 16}}{{25}}} \]
On calculating
\[{e_1} = \dfrac{3}{5}\]
We have given that \[{e_1} \times {e_2} = 1\]
On putting the value of \[{e_1}\] we are able to find the value of \[{e_2}\]
\[\dfrac{3}{5} \times {e_2} = 1\]
On rearranging we got the value of the eccentricity of the hyperbola-
\[{e_2} = \dfrac{5}{3}\]
Formula of eccentricity of the hyperbola \[e = \sqrt {1 + \dfrac{{{a^2}}}{{{b^2}}}} \]
On putting the values.
\[\dfrac{5}{3} = \sqrt {1 + \dfrac{{{a^2}}}{{{b^2}}}} \]
Squaring both side.
\[\dfrac{{25}}{9} = 1 + \dfrac{{{a^2}}}{{{b^2}}}\]
On further calculations
\[\dfrac{{{a^2}}}{{{b^2}}} = \dfrac{{16}}{9}\] ……(i)
The hyperbola is passing through the foci of the ellipse.
The coordinate of the ellipse are \[\left( {0, \pm 3} \right)\]
Form here value is \[{b^2} = 9\]
On putting this value in equation (i)
\[\dfrac{{{a^2}}}{9} = \dfrac{{16}}{9}\]
On further calculation.
\[{a^2} = 16\]
The required hyperbola satisfying all the conditions given in question is
\[\dfrac{{{x^2}}}{{16}} - \dfrac{{{y^2}}}{9} = - 1\]
Final answer:
According to the obtained answer option b is the correct answer.

Note: In this type of question, students are usually confused in the formula of the eccentricity of the ellipse and the hyperbola. They exchange the formula of the eccentricity of both the ellipse and the hyperbola. This question can be done by another method that is checking all the options.