Question

If 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, then the value of $b^2$ is [MNR 1992; UPSEAT 2001; AIEEE 2003; Karnataka CET 2004; Kerala (Engg.) 2005]
(A) 1
(B) 5
(C) 7
(D) 9

Answer 1

Hint: In the given question, we are told that the foci of the hyperbola and ellipse coincide. This means that their focus is the same. We will first calculate the eccentricity of the hyperbola, from which the focus of the hyperbola is deduced. This focus point is also the focus of the ellipse. So we will calculate the eccentricity of the ellipse and then the value of $b^2$.



Complete step by step solution:The given equation of hyperbola is $\dfrac{x^{2}}{144}-\dfrac{y^{2}}{81}=\dfrac{1}{25}$
Rewriting this equation, we get $\dfrac{x^{2}}{\dfrac{144}{25}}-\dfrac{y^{2}}{\dfrac{81}{25}}=1$
From here we get, a = $\\surd\dfrac{144}{25}$ = $\dfrac{12}{5}$ and b = $\\surd\dfrac{81}{25}$ = $\dfrac{9}{5}$
The eccentricity of the hyperbola is calculated as $e= \dfrac{\surd a^{2}+b^{2}}{a}$.
The focus of the hyperbola is (c, 0).
Also, as $e=\dfrac{c}{a}$. So, the c can be written as c = ae.
The focus of the hyperbola is, thus, (ae,0) = ($\dfrac{12}{5}.\dfrac{5}{4}$,0) = (3,0)
As the foci of an ellipse and a hyperbola coincide, this means their focus is at the same point. Thus, the focus of the ellipse is also (3, 0).
The given equation of the ellipse is $\dfrac{x^{2}}{16}+\dfrac{y^{2}}{b^{2}}=1$.
From here, we can see that a = 4. The focus of the ellipse is given by (ae,0).
Hence, (ae,0) = (3,0)
(4e,0) = (3,0)
e = $\dfrac{3}{4}$
The eccentricity of the ellipse is given by $e= \dfrac{\surd a^{2}-b^{2}}{a}$.
$\dfrac{3}{4}= \dfrac{\surd 4^{2}-b^{2}}{4}$
$9=16-b^{2}$
$b^2=7$
Thus, If 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, then the value of $b^2$ is 7.



Option ‘C’ is correct

Note: In a hyperbola, there are two axes. The transverse axis is the line that passes through the foci. The conjugate axis is the line through the centre that is also perpendicular to the transverse axis. An ellipse contains two axes. One is the major axis, which passes through the foci of the ellipse, and another is the minor axis.