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The foci of the ellipse $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ and the hyperbola $\dfrac{{{x^2}}}{{144}} - \dfrac{{{y^2}}}{{81}} = \dfrac{1}{{25}}$ coincide, the value of ${b^2}$ is
A. 1.
B. 5.
C. 7.
D. 9.

Last updated date: 17th Jun 2024
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Hint: To solve the question, we have to remember that, in terms of eccentricity, the focus is at the distance of ae from the centre or we can say that the foci will be (ae, 0). And if foci of ellipse and hyperbola coincide, then foci of hyperbola = foci of ellipse.
We know that, the eccentricity of an ellipse and hyperbola is the ratio from the centre to one of the foci and to one of the vertices and is given by $e = \dfrac{{\sqrt {{a^2} - {b^2}} }}{a}$ for ellipse and $e = \dfrac{{\sqrt {{a^2} + {b^2}} }}{a}$ for hyperbola.

Complete step-by-step answer:
Given that,
Equation of ellipse = $\dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$
Equation of hyperbola = $\dfrac{{{x^2}}}{{144}} - \dfrac{{{y^2}}}{{81}} = \dfrac{1}{{25}}$
It can also be written as:
$\dfrac{{25{x^2}}}{{144}} - \dfrac{{25{y^2}}}{{81}} = 1$, or
$\dfrac{{{x^2}}}{{\dfrac{{144}}{{25}}}} - \dfrac{{{y^2}}}{{\dfrac{{81}}{{25}}}} = 1$
Now, we know that the eccentricity of ellipse is given by, $e = \dfrac{{\sqrt {{a^2} - {b^2}} }}{a}$
Comparing the equation of ellipse with general equation of ellipse i.e. $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$
We get a = 4
The eccentricity of ellipse will be,
$ \Rightarrow e = \dfrac{{\sqrt {16 - {b^2}} }}{4}$
$ \Rightarrow e = \sqrt {1 - \dfrac{{{b^2}}}{{16}}} $ ………. (i)
Now, comparing the equation of hyperbola with general equation of hyperbola i.e. $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$
We get, $a' = \dfrac{{12}}{5}$ and $b' = \dfrac{9}{5}$
Then, the eccentricity of hyperbola will be,
$ \Rightarrow e' = \dfrac{{\sqrt {\dfrac{{144}}{{25}} + \dfrac{{81}}{{25}}} }}{{\dfrac{{12}}{5}}}$
$ \Rightarrow e' = \dfrac{3}{{\dfrac{{12}}{5}}}$ ……… (ii)
Now, we know that the hyperbola and ellipse coincide, then their foci will be equal.
i.e. foci of ellipse = foci of hyperbola.
$ \Rightarrow ae = a'e'$
Putting the values from equation (i) and (ii),
$ \Rightarrow 4\sqrt {1 - \dfrac{{{b^2}}}{{16}}} = \dfrac{{12}}{5} \times 3 \times \dfrac{5}{{12}}$
$ \Rightarrow \sqrt {1 - \dfrac{{{b^2}}}{{16}}} = \dfrac{3}{4}$
Squaring both sides, we will get
$ \Rightarrow \left( {1 - \dfrac{{{b^2}}}{{16}}} \right) = \dfrac{9}{{16}}$
Taking L.C.M, we get
\[ \Rightarrow 16 - {b^2} = \dfrac{9}{{16}} \times 16\]
\[ \Rightarrow 16 - {b^2} = 9\]
\[ \Rightarrow {b^2} = 16 - 9\]
\[ \Rightarrow {b^2} = 7\]

So, the correct answer is “Option C”.

Note: The foci of an ellipse always lie on the major axis. The major axis is determined by finding the intercepts on the axes of symmetry. Whereas, the foci of a hyperbola always lie on the transverse axis. It is the positive term whose denominator gives the transverse axis.