A) \[{{x}^{2}}\cos e{{c}^{2}}\theta -{{y}^{2}}{{\sec }^{2}}\theta =1\]
B) \[{{x}^{2}}{{\sec }^{2}}\theta -{{y}^{2}}\cos e{{c}^{2}}\theta =1\]
C) \[{{x}^{2}}{{\sin }^{2}}\theta -{{y}^{2}}{{\cos }^{2}}\theta =1\]
D) \[{{x}^{2}}{{\cos }^{2}}\theta -{{y}^{2}}{{\sin }^{2}}\theta =1\]
Correct Answer: A
Solution :
| [a] Equation of the ellipse is \[3{{x}^{2}}+4{{y}^{2}}=12\] |
| \[\Rightarrow \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{3}=1\] ?. (1) |
| Eccentricity \[{{e}_{1}}=\sqrt{1-\frac{3}{4}}=\frac{1}{2}\] |
| So, the foci of ellipse are \[(1,0)\] and \[(-1,0)\] |
| Let the equation of the required hyperbola be |
| \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] ?. (2) |
| Given \[2a=2\sin \theta \Rightarrow a=\sin \theta \] |
| Since the ellipse (1) and the hyperbola (2) are confocal, so. |
| The foci of hyperbola are \[(1,0)\] and \[(-1,0)\]too. It the eccentricity, of hyperbola be \[{{e}_{2}}\] then \[a{{e}_{2}}=1\Rightarrow \sin \,\theta {{e}_{2}}=1\Rightarrow {{e}_{2}}=\cos ec\,{{\theta }^{2}}\] |
| \[\therefore {{b}^{2}}={{a}^{2}}({{e}^{2}}_{2}-1)=si{{n}^{2}}\theta (cose{{c}^{2}}\theta -1)=co{{s}^{2}}\theta \] |
| \[\therefore \] Required equation of the hyperbola is |
| \[\frac{{{x}^{2}}}{{{\sin }^{2}}\theta }-\frac{{{y}^{2}}}{{{\cos }^{2}}\theta }=1\Rightarrow {{x}^{2}}\cos e{{c}^{2}}\theta -{{y}^{2}}{{\sec }^{2}}\theta =1\] |
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