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. \(\rm \frac {y^2} 7 + \frac {x^2} 9 = 1\)
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. \(\rm \frac {y^2} 7 + \frac {x^2} 9 = 1\)
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 \(\rm \frac {y^2} 7 – \frac {x^2} 9 = 1\)