Question

If the equation $\sqrt{(x+4{)}^2+(y-3{)}^2}-\sqrt{(x-4{)}^2+(y-3{)}^2}=6$ represents a conic its directrices are :

• A. $x=\pm \frac{9}{4}$
• B. $x=\pm \frac{9}{2}$
• C. $x=\pm \frac{7}{3}$
• D. $x=\pm \frac{7}{2}$

Answer 1

Answer: (A)

$x=\pm \frac{9}{4}$

$\sqrt{(x+4{)}^2+(y-3{)}^2}-\sqrt{(x-4{)}^2+(y-3{)}^2}=6$
Clearly this is of the form, $|SP-S^{\prime }P|=2a$
Thus $S\equiv (-4,3),S^{\prime }\equiv (4,3)$
Also $2a=6\Rightarrow a=3$
$\Rightarrow ae=4\Rightarrow e=\frac{4}{3}$
Thus directrix of the given hyperbola is $x=\pm \frac{a}{e}=\pm \frac{9}{4}$
Hence, option 'A' is correct.