Question

Find the equation of the hyperbola whose foci are s(6 4) and s'(-4 4) and eccentricity 2

Answer 1

Please solve it i hint,
Because,2ae=6+4=10
e=2its given
And, b^2=3a^2
Thus ,a=5/2&b=5√3/2

Answer 2

Therefore the equation of hyperbola is

$12{(x- 1)^2}- 4{(y- 4)^2}=75$

Step-by-step explanation:

The co-ordinates of foci of a hyperbola are $(\alpha+ae,\beta)$  and  $(\alpha-ae,\beta)$

where the center of the hyperbola is $(\alpha,\beta)$

e= eccentricity $=\sqrt{1+\frac{b^2}{a^2}}$

a= semi major axis.

Then the equation of hyperbola

$\frac{(x-\alpha)^2}{a^2}-\frac{(y-\beta)^2}{b^2}=1$

Given the the co-ordinate of foci are (6,4) and (-4,4). Eccentricity = e= 2

$(\alpha+ae,\beta)=(6,4)$  and   $(\alpha-ae,\beta)=(-4,4)$

$\therefore \beta=4$

$\alpha+ae=6$ ......(1)   and       $\alpha-ae=-4$......(2)

Adding equation (1) and (2)

$\therefore\alpha+ae+\alpha-ae=6+(-4)$

$\Rightarrow 2\alpha =2$

$\Rightarrow \alpha =1$

Now putting $\alpha =1 \ and \ e=2$ in equation (1)

$1+a.2=6$

$\Rightarrow a=\frac52$

We know that

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

$\Rightarrow b^2= a^2(e^2-1)$

$\Rightarrow b^2= (\frac52)^2(2^2-1)$

$\Rightarrow b^2= \frac{75}{4}$

Therefore the equation of hyperbola is

$\frac{(x-\alpha)^2}{a^2}-\frac{(y-\beta)^2}{b^2}=1$

$\frac{(x- 1)^2}{\frac{2 5}{4}}- \frac{(y- 4)^2}{\frac {75}{4}}=1$

$12{(x- 1)^2}- 4{(y- 4)^2}=75$