Question

The equation of hyperbola whose coordinates of the foci are $(\pm 8,0)$ and the lenght of latus rectum is $24$ units, is

• A. $3x^2-y^2=48$
• B. $4x^2-y^2=48$
• C. $x^2-3y^2=48$
• D. $x^2-4y^2=48$

Answer 1

Answer: (A)

$3x^2-y^2=48$

The Foci of hyperbola are $(\pm 8,0)$, hence Foci lie on $x$ - axis.

We know that foci of hyperbola lie at $(\pm ae,0)$, So $ae=8$ ...$\left(1\right)$

squaring both sides of equation $\left(1\right)$, we get,

$\Rightarrow a^2{e}^2=64$

Eccentricity of hyperbola $e^2=1+\frac{b^2}{a^2}$

$\Rightarrow a^2(1+\frac{b^2}{a^2})=64$

$\Rightarrow a^2+b^2=64$ ...$\left(2\right)$

Now the length of latus rectum is given as 24 units.

length of latusrectum of hyperbola $=\frac{2b^2}{a}=24$

$\Rightarrow b^2=12a$ ...$\left(3\right)$

putting value of $b^2$ in eq. $\left(2\right)$, we get,

$\Rightarrow a^2+12a-64=0$

Hence $a=4,-16$

As $a$ is always taken as positive value so $a=4$

from eq. $\left(3\right)$, $b=\sqrt{48}$

Hence equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{48}=1$

Or $3x^2-y^2=48$, So correct option is $A$.