Question

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

• A. $x-1=0$
• B. $x-1\pm \frac{3}{\sqrt{19}}=0$
• C. $x+1\pm \frac{3}{\sqrt{19}}=0$
• D. $x-1\pm \frac{1}{\sqrt{19}}=0$

Answer 1

The correct option is B $x-1\pm \frac{3}{\sqrt{19}}=0$

Comparing the equation of given hyperbola with the standard equation

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

$h=1,k=-2,a^2=16,b^2=3$

$e=\sqrt{1+\frac{b^2}{a^2}}=\frac{\sqrt{19}}{4}$

Equation of directrix is $y-k=\pm \frac{a}{e}$

$⟹y+2=\pm \frac{16}{\sqrt{19}}$