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Question: The centre, one vertex, one focus of a hyperbola are (5, 4), (5, 7), (5, 9). Its directrices are: ...

The centre, one vertex, one focus of a hyperbola are (5, 4), (5, 7), (5, 9). Its directrices are:
A) 5y – 27 = 0, 5y – 9 = 0
B) 5y – 17 = 0, 5y – 11 = 0
C) 5y – 29 = 0, 5y – 11 = 0
D) 5y – 7 = 0, 5y – 17 = 0

Explanation

Solution

We will first find out the values of a, which we can denote as the distance between the centre of the hyperbola and the vertex of the hyperbola, and ae, the distance between the centre and the focus using the distance formula: distance = (x2−x1)2+(y2−y1)2\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} . And from these two, we will calculate the eccentricity e of the hyperbola. The equation of the directrix is given as: y – 4 = ±ae \pm \dfrac{a}{e} since the axis of the hyperbola is parallel to y – axis. Upon substituting the values of a and e, we will get the equation of the directrices of the hyperbola.

Complete step by step solution:
We are given the centre of hyperbola as (5, 4), one vertex of hyperbola has co-ordinates (5, 7) and one of the focus is at (5, 9).
We can calculate the distance between the centre of the hyperbola and the vertex of the hyperbola, a, using the distance formula: (x2−x1)2+(y2−y1)2\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}
⇒\Rightarrowa = (5−5)2+(7−4)2=0+9=9=3\sqrt {{{\left( {5 - 5} \right)}^2} + {{\left( {7 - 4} \right)}^2}} = \sqrt {0 + 9} = \sqrt 9 = 3
Therefore, a = 3.
Similarly, we can calculate the distance between the centre of the hyperbola and the focus
of the hyperbola, ae, using the distance formula: (x2−x1)2+(y2−y1)2\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}
⇒ae=(5−5)2+(9−4)2\Rightarrow ae = \sqrt {{{\left( {5 - 5} \right)}^2} + {{\left( {9 - 4} \right)}^2}} = 0+25=25=5\sqrt {0 + 25} = \sqrt {25} = 5
Therefore, ae = 5
Now, we can calculate the eccentricity of the given hyperbola using ae and a as: e = aea\dfrac{{ae}}{a}
⇒e=53\Rightarrow e = \dfrac{5}{3}
Therefore, the equation of the directrices of the hyperbola will be given by: y – 4 = ±ae \pm \dfrac{a}{e}as the axis of the hyperbola is parallel to y – axis.
Putting the values of a and e in this equation, we get
⇒y−4=±353=±95\Rightarrow y - 4 = \pm \dfrac{3}{{\dfrac{5}{3}}} = \pm \dfrac{9}{5}
Or, we can write this as
⇒y=4±95\Rightarrow y = 4 \pm \dfrac{9}{5}
Therefore, y can have two values i.e., y=4+95y = 4 + \dfrac{9}{5} and y=4−95y = 4 - \dfrac{9}{5}
Hence, we can write these equation in simplified form as:
⇒5y=29 and 5y=11\Rightarrow 5y = 29{\text{ and 5}}y = 11
Or, 5y−29=0 and 5y−11=05y - 29 = 0{\text{ and 5}}y - 11 = 0

Hence, option (C) is correct.

Note:
You may go wrong while using the equation of the directrix of the hyperbola because here, we have used the y – coordinate as the axis of the hyperbola was parallel to y – axis (since x – coefficient of centre, vertex and focus is constant).