CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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

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 proved

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image