Question

For hyperbola $-\frac{(x-1{)}^2}{3}+\frac{(y+2{)}^2}{16}=1$, vertices are

• A. $(2,3)$
• B. $(\pm \sqrt{3},3)$
• C. $(2,\pm 3)$
• D. $(1,2)$,$(1,-6)$

Answer 1

The correct option is D $(1,2)$,$(1,-6)$

Hyperbola $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

It has center at $(0,0)$ and vertex at $(0,\pm a)$.

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

Its center shifted to $(1,-2)$.

So, vertex will also shift by $(1,-2)$.

So, vertex is $(0,\pm a)+(1,-2)=(0,\pm 4)+(1,-2)$

$\Rightarrow (1,2)$ and $(1,-6)$