Question

He equations of the directrices of the conic

$x^{2} + 2x - y^{2} + 5 = 0$ are

• A. $x = \pm 1$
• B. $y = \pm 2$
• C. $y = \pm \sqrt{2}$
• D. $x = \pm \sqrt{3}$

Answer 1

Answer: (C)

$y = \pm \sqrt{2}$

Answer 2

$(x + 1)^{2} - y^{2} - 1 + 5 = 0$ ⇒ $a^{2}l^{2}\tan^{2}\varphi + b^{2}m^{2}\sec^{2}\varphi + 2ablm\sec\varphi\tan\varphi = n^{2}$

Equation of directrices of $\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$ are $y = \pm \frac{b}{e}$

Here $b = 2$, $al\tan\varphi + bm\sec\varphi = - n$. Hence, $y = \pm \frac{2}{\sqrt{2}}$ ⇒ $y = \pm \sqrt{2}$