Question

Find the equation of hyperbola if the eccentricity ${e_1}$ of the ellipse is $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{{25}} = 1$ and ${e_2}$ is the eccentricity of the hyperbola respectively which passes through the foci of an ellipse given that ${e_1}{e_2} = 1$ satisfies the equations.
(a) $\dfrac{{{x^2}}}{{16}} - \dfrac{{{y^2}}}{9} = - 1$
(b) $\dfrac{{{x^2}}}{{16}} - \dfrac{{{y^2}}}{9} = 1$
(c) $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = - 1$
(d) None of these

Answer 1

Hint : The given problem revolves around the concepts of curved equations like parabola, hyperbola, eclipse, etc. So, we will first analyze the given equation with generalised formula for ellipse $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{{25}} = 1$ as well as hyperbola $\dfrac{{{y^2}}}{{{b^2}}} - \dfrac{{{x^2}}}{{{a^2}}} = 1$ particularly. Then, by calculating foci and using the given conditions, substituting the values in the required equation, the desired solution can be obtained.

Complete step-by-step answer :
Since, we have given the equation for the ellipse respectively,
$\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{{25}} = 1$
$\because$ We know that,
The generalized formula or an equation to represent ellipse (in the form of eccentricity) is,
$\Rightarrow \dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$
Where, $a$ and $b$ passes through the ellipse respectively
Comparing the given equation to the above standardized equation of ellipse, we get
$\therefore a = 4,b = 5$
Since, we have given that
Any eccentricity $e = \sqrt {1 - \dfrac{{{a^2}}}{{{b^2}}}}$ defines as,
As a result, substituting the values to get the eccentricity ${e_1}$ of an ellipse respectively
${e_1} = \sqrt {1 - \dfrac{{16}}{{25}}}$
Solving the equation mathematically, we get
${e_1} = \sqrt {\dfrac{{25 - 16}}{{25}}} = \sqrt {\dfrac{9}{{25}}} \\\ {e_1} = \dfrac{3}{5} = \dfrac{b}{a} \\\$
$\therefore a = 5,b = 3$
Also,
We have given,
‘ ${e_2}$ ’ is an eccentricity hyperbola,
As a result, by the second condition that is ${e_1}{e_2} = 1$ , it seems that
Substituting ${e_1} = \dfrac{3}{5}$ , we get
$\left( {\dfrac{3}{5}} \right){e_2} = 1$
Simplifying the equation mathematically, we get
${e_2} = \dfrac{5}{3}$
Also, we know that
Foci of an ellipse are $= \left( {0, \pm be} \right)$
Foci of ellipse is $= \left( {0, \pm 3} \right)$
Equation of hyperbola can be represented by $\dfrac{{{y^2}}}{{{b^2}}} - \dfrac{{{x^2}}}{{{a^2}}} = 1$ respectively,
But, we have given that ${e_2}$ passes through the focii of an ellipse,
Hence, the equation satisfies the given elliptical equation with hyperbolic parameters, as a result we get
Hence, we get
$\therefore {b^2} = {\left( { \pm 3} \right)^2} = 9$ also,
${a^2} = {b^2}\left( {{e^2}_2 - 1} \right) = 9\left( {\dfrac{{25}}{9} - 1} \right) \\\ \therefore {a^2} = 9 \times \dfrac{{16}}{9} = 16 \\\$
The required equation of hyperbola is,
(by substituting these values, we get)
$\dfrac{{{y^2}}}{9} - \dfrac{{{x^2}}}{{16}} = 1$
This equation can also be rearranged as,
$\dfrac{{{x^2}}}{{16}} - \dfrac{{{y^2}}}{9} = - 1$ respectively
$\Rightarrow \therefore$ The option (a) is correct!
So, the correct answer is “Option a”.

Note : One must know able to compare the given equation of curve with respect to curve being asked to solve, here the curve is ellipse and hyperbola been asked which is represented by $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{{25}} = 1$ & $\dfrac{{{y^2}}}{{{b^2}}} - \dfrac{{{x^2}}}{{{a^2}}} = 1$ and its respective parameter like foci $\left( {0, \pm be} \right)$ , etc. Remember eccentricities of each curves $e = \sqrt {1 - \dfrac{{{a^2}}}{{{b^2}}}}$ to find any eccentricity of the curve, so as to be sure of our final answer.