InterviewSolution
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InterviewSolution
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The equation of the hyperbola with center at the origin, length of the transverse axis is 6 and one focus at (0, 4) is ?1. \(\rm \frac {y^2} 7 - \frac {x^2} 9 = 1\)2. \(\rm \frac {y^2} 9 - \frac {x^2} 7 = 1\)3. \(\rm \frac {y^2} 9 + \frac {x^2} 9 = 1\)4. \(\rm \frac {y^2} 7 + \frac {x^2} 9 = 1\) |
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Answer» Correct Answer - Option 1 : \(\rm \frac {y^2} 7 - \frac {x^2} 9 = 1\) Concept: The equation of the hyperbola is \(\rm \dfrac {y^2}{b^2}- \dfrac{x^2}{a^2} = 1\) with the foci (0 , ± ae) Length of the transverse axis = 2a
Calculations: Since the coordinates of the one focus at (0, 4) = (0 , ± ae) , it is a case of vertical hyperbola ⇒ ae = 4 It is a case of vertical hyperbola ⇒ The equation of hyperbola is \(\rm \dfrac {y^2}{b^2}- \dfrac{x^2}{a^2} = 1\) ....(1) Length of the transverse axis = 6 ⇒ 2a=6 ⇒ a = 3 \(\rm \text {Also}\;\;a^2e^2 = a ^ 2+ b^2\) ⇒\(\rm 4^2 =3 ^ 2+ b^2\) ⇒\(\rm b^2 = 7\) Equation (1) becomes \(\rm \frac {y^2} 7 - \frac {x^2} 9 = 1\) Hence, The equation of the hyperbola with center at the origin, length of the transverse axis is 6 and one focus at (0, 4) is \(\rm \frac {y^2} 7 - \frac {x^2} 9 = 1\) |
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