Question

For hyperbola $-\frac{(x-1{)}^2}{3}+\frac{(y+2{)}^2}{16}=1$ centre is

• A. $(-1,-2)$
• B. $(1,-1)$
• C. $(1,-2)$
• D. $(0,0)$

Answer 1

Answer: (C)

$(1,-2)$

The given hyperbola is $-\frac{(x-1{)}^2}{3}+\frac{(y+2{)}^2}{16}=1$ is a conjugate hyperbola.

It can be written as $\frac{(x-1{)}^2}{3}-\frac{(y+2{)}^2}{16}=-1$ ......$\left(1\right)$

Now let $x-1=X$ and $y+2=Y$

Putting values of $x-1$ and $y+2$ in eq. $\left(2\right)$ we get,

$\to \frac{(X{)}^2}{3}-\frac{(Y{)}^2}{16}=-1$ ....$\left(2\right)$

We can see the eq$\left(2\right)$ is a standard form of conjugate hyperbola and it's center lies at origin $(0,0)$

So $X=0,Y=0$ is the center of the hyperbola given in eq.$\left(2\right)$

$\to$ $X=x-1=0$ or $x=1$

$\to$ $Y=y+2=0$ or $y=-2$

Hence center of the given hyperbola in eq. $\left(1\right)$ is $(1,-2)$. So correct option is $C$.