Harshit Singh
Last Activity: 4 Years ago
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Given that: Foci (0, ±13), Conjugate axis length = 24
It is noted that the foci are on the y-axis.
Therefore, the equation of the hyperbola is of the form:
(y^2/a^2)-(x^2/b^2) = 1 …(1)
Since the foci are (0, ±13), we can get
C = 13
It is given that, the length of the conjugate axis is 24,
It becomes 2b = 24
b= 24/2
b= 12
And, we know that a^2+ b^2= c2
To find a, substitute the value of b and c in the above equation:
a^2+ 122= 132
a^2= 169-144
a^2= 25
Now, substitute the value of a and b in equation (1), we get
(y^2/25)-(x^2/144) = 1, which is the required equation of the hyperbola.
Thanks