Conic Sections Question 454
Question: The equation of the directrices of the conic $ x^{2}+2x-y^{2}+5=0 $ are
Options:
A) $ x=\pm 1 $
B) $ y=\pm 2 $
C) $ y=\pm \sqrt{2} $
D) $ x=\pm \sqrt{3} $
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Answer:
Correct Answer: C
Solution:
$ {{(x+1)}^{2}}-y^{2}-1+5=0 $
therefore $ -\frac{{{(x+1)}^{2}}}{4}+\frac{y^{2}}{4}=1 $
Equation of directrices of $ \frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1 $ are $ y=\pm \frac{b}{e} $
Here $ b=2,\ e=\sqrt{1+1}=\sqrt{2} $
Hence $ y=\pm \frac{2}{\sqrt{2}} $
therefore $ y=\pm \sqrt{2} $ .
BETA
