Question
Question: How do you find the standard form of the equation of the hyperbola given the properties foci \(( \pm...
How do you find the standard form of the equation of the hyperbola given the properties foci (±5,0) , length of the conjugate axis 6 ?
Solution
For solving this particular question , we first define the general form of equation , then we compare the given foci with the general form of foci , we know that the length of the conjugate axis is given by 2b , substituting all we will get the desired result.
Complete step by step solution:
We have the foci (±5,0) , therefore we can say that the hyperbola is the horizontal transverse axis type hyperbola ,
The general equation is given by ,
a2(x−h)2−b2(y−k)2=1......(1)
Now, we have the length of the conjugate axis that is given by 2b=6,
Therefore, we get b=3..............(2)
Now, we also know that the general forms of foci is given by ,
(h−a2+b2,k) and (h+a2+b2,k)
By equating general form with the given foci that is (−5,0) and (5,0).
We will get ,
k=0....................(3) ,
h−a2+b2=−5....................(4),
h+a2+b2=5......................(5) .
Now adding (4)and (5) , we will get ,
h=0..................(6)
Now, substitute (2) and (6) in equation (5) ,
We will get ,
⇒h+a2+b2=5
⇒0+a2+32=5
Taking square both the side ,
⇒a2+32=52
Subtract 32 from both the side ,
⇒a2=52−32
⇒a2=42
Now, taking square root both the side,
⇒a=4......................(7)
Now, substitute (2),(3),(6) and (7) in equation (1) ,
We will get ,
⇒a2(x−h)2−b2(y−k)2=1 ⇒42(x−0)2−32(y−0)2=1
Hence we get the required equation .
Note: The important thing to recollect about any equation is that the ‘equals’ sign represents a balance. What the sign says is that what’s on the left-hand side is strictly an equal to what’s on the right-hand side.