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Information : The given hyperbola is a horizontal hylerbola with its centre (3 , -5) and one of its focus at (9 , -5) and vertex at (7 , -5) and as we can see that the focus and vertex have same y - coordinates, it must have its Transverse axis on line y = - 5.
Now,
it's vertex is given, I.e (7 , -5)
so, length of semi transverse axis will be equal to distance of vertex from centre, i.e
Now, it's focus can be represented as ;

so,
and we know, a = 4



Now, let's find the measure of semi - conjugate axis (b)






So, it's time to write the equation of hyperbola, as we already have the values of a and b ~

[ plug in the values, and h = x - coordinate of centre, and k = y - coordinate of centre ]

Step-by-step explanation:
Find the GCD (or HCF) of numerator and denominator
GCD of 41 and 15 is 1
Divide both the numerator and denominator by the GCD
41 ÷ 1
15 ÷ 1
Reduced fraction:
41
15
Therefore, 41/15 simplified to lowest terms is 41/15.
Answer:
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Answer:
Substitute the slope and the coordinates of point P in y = mx + b and then solve for b for each equation.
Step-by-step explanation: