Question

The equation of the hyperbola whose centre is(6,2) one focus is (4,2) and of eccenticity 2 is

• A. $3(x-6{)}^2-(y-2{)}^2=3$
• B. $(x-6{)}^2-3(y-2{)}^2=1$
• C. $(x-6{)}^2-2(y-2{)}^2=1$
• D. $2(x-6{)}^2-(y-2{)}^2=3$

Answer 1

The correct option is A

$3(x-6{)}^2-(y-2{)}^2=3$

$\text{The equation of the hyperbola with centre}(x_0,y_0)isgivenby$

$\frac{(x_0,x_0{)}^2}{a^2}-\frac{(y_0,y_0)}{b^2}=1$

$Focus=(ae+x_0,y_0)$

$\therefore ae=-2$

$\Rightarrow a=-1$

$b^2=(2{)}^2-a^2$

$\Rightarrow b^2=(-2{)}^2-(-1{)}^2$

$\Rightarrow b^2=3$

$\therefore \frac{(x-6{)}^2}{1}-\frac{(y-2{)}^2}{3}=1$

$\Rightarrow 3(x-6{)}^2-(y-2{)}^2=3$