Question

The length of the transverse axis of a hyperbola is\[2\cos (\alpha )\]. The foci of the hyperbola are the same as that of the ellipse\[9{{x}^{2}}+16{{y}^{2}}=144\]. The equation of the hyperbola is
A. \[\dfrac{{{x}^{2}}}{{{\cos }^{2}}\alpha }-\dfrac{{{y}^{2}}}{7-{{\cos }^{2}}\alpha }=1\]
B. \[\dfrac{{{x}^{2}}}{{{\cos }^{2}}\alpha }-\dfrac{{{y}^{2}}}{7+{{\cos }^{2}}\alpha }=1\]
C. \[\dfrac{{{x}^{2}}}{1+{{\cos }^{2}}\alpha }-\dfrac{{{y}^{2}}}{7-{{\cos }^{2}}\alpha }=1\]
D. \[\dfrac{{{x}^{2}}}{{{\cos }^{2}}\alpha }-\dfrac{{{y}^{2}}}{7-{{\cos }^{2}}\alpha }=1\]

Answer 1

Hint: The given equation \[9{{x}^{2}}+16{{y}^{2}}=144\] is written in the form of\[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]. Foci of the ellipse is the same as the foci of hyperbola. But the foci of the ellipse is\[(\pm ae,0)\]. Transverse axis of the hyperbola is \[2a\] .That means the axis of the hyperbola between two foci. The length segment between two foci is 2a.

Complete step by step answer:
The given equation is \[9{{x}^{2}}+16{{y}^{2}}=144\] can be written in the form of
\[\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\,\,---(1)\]
That is\[\dfrac{{{x}^{2}}}{16}+\dfrac{{{y}^{2}}}{9}=1\,\,\].
\[\therefore \dfrac{{{x}^{2}}}{{{4}^{2}}}+\dfrac{{{y}^{2}}}{{{3}^{2}}}=1\,\,---(2)\]
By comparing the equations \[(1)\]and\[(2)\] we will get:
\[a=4\]
 \[b=3\]
To find the coordinates of foci of ellipse that is\[(\pm ae,0)\].
But, the value of \[a=4\]
To find value of \[e\]is:
Formula for \[e\] is:
\[e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}\]
After substituting the values of \[a\]and\[b\]we get:
\[e=\sqrt{1-\dfrac{9}{16}}\]
Further simplifying we get:
\[e=\sqrt{\dfrac{7}{16}}\]
After simplifying further we will get:
\[e=\dfrac{\sqrt{7}}{4}\]
Foci of the ellipse is \[(\pm ae,0)\]
\[\left( \pm 4\times \dfrac{\sqrt{7}}{4},0 \right)\Rightarrow \left( \pm \sqrt{7},0 \right)\,\,---(3)\]
Foci of ellipse is same as the foci of hyperbola
\[\therefore \] Foci of hyperbola is \[\left( \pm \sqrt{7},0 \right)\]
Transverse axis of the hyperbola is \[2a\] because the transverse axis is the axis of the hyperbola between two foci. The line segment between two foci is\[2a\].
By comparing \[2a\] with given the length of the transverse axis of a hyperbola is \[2\cos (\alpha )\]
\[2a=2\cos (\alpha )\]
By cancelling 2 on this equation we get:
\[a=\cos \alpha ----(4)\]
By comparing the equation \[(3)\]with Foci of the ellipse that is \[(\pm ae,0)\]you will get:
\[ae=\sqrt{7}----(5)\]
By substituting the value of equation \[(4)\] in equation\[(5)\]you will get:
\[(\cos \alpha )\times e=\sqrt{7}\]
\[e=\dfrac{\sqrt{7}}{\cos \alpha }----(6)\]
For hyperbola,
\[{{e}^{2}}=1+\dfrac{{{b}^{2}}}{{{a}^{2}}}----(7)\]
By substituting the values of equation \[(4)\] and equation \[(6)\]in equation \[(7)\]
\[\dfrac{7}{{{\cos }^{2}}\alpha }=1+\dfrac{{{b}^{2}}}{{{\cos }^{2}}\alpha }\]
Further simplifying we get:
\[\dfrac{7}{{{\cos }^{2}}\alpha }=\dfrac{{{\cos }^{2}}\alpha +{{b}^{2}}}{{{\cos }^{2}}\alpha }\]
\[\therefore {{b}^{2}}=7-{{\cos }^{2}}\alpha \]
So, the equation of hyperbola is
\[\dfrac{{{x}^{2}}}{{{\cos }^{2}}\alpha }-\dfrac{{{y}^{2}}}{7-{{\cos }^{2}}\alpha }=1\]

So, the correct answer is “Option A”.

Note: Transverse axis of hyperbola is the axis of the hyperbola between the two foci. Line segment between two foci is called the conjugate axis. Hence the transverse axis of hyperbola is\[2a\]. Remember that foci of the ellipse is\[(\pm ae,0)\].