Question

# The vertices of a hyperbola are at $$( 0,0 )$$ and $$( 10,0 )$$ and one of its foci is at $$( 18,0 )$$ . The equation of the hyperbola is

A
x225y2144=1
B
(x5)225y2144=1
C
x225(y5)2144=1
D
(x5)225(y5)2144=1

Solution

## The correct option is B $$\displaystyle \frac { ( x - 5 ) ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 144 } = 1$$vertices are $$A(0,0)$$ and $$A'(10,0)$$$$AA' = 2a$$$$\Rightarrow 10 = 2a$$$$\Rightarrow a = 5$$Focus is$$ae-a = 8$$$$\Rightarrow e - 1 = \dfrac{8}{5}$$$$\Rightarrow e = 1 + \dfrac{8}{5} = \dfrac{13}{5}$$Now, $$b = a\sqrt{e^2-1}$$$$b = 5\sqrt{\dfrac{169}{25}-1}$$$$= 5\dfrac{12}{5} = 12$$Since, centre $$=(ae,0)=(5,0)$$$$\therefore$$ Required equation $$\dfrac{(x-5)^2}{25} - \dfrac{y^2}{144} - 1$$Hence provedMaths

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