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The diretrix of the parabola $y^{2}=x+4$ is

\[\begin{array}{1 1}(1)x=\frac{15}{4}&(2)x=-\frac{15}{4}\\(3)\frac{17}{4}&(4)x=-\frac{17}{4}\end{array}\]

1 Answer

$y^2=x+4$
$(y-0)^2=(x+4)$
$y^2=X$ Where $Y= y-0 \;X=x+4$
$Y^2= 4 \bigg( \large\frac{1}{4} \bigg) $$x$
$a= \large\frac{1}{4}$
The equation of the directrix is
$x= -\large\frac{1}{4}$
$x+4=-\large\frac{1}{4}$
$x= -\large\frac{1}{4}$$-4 $
$\quad= \large\frac{-1-16}{4}$
$\quad= \large\frac{-17}{4}$
Hence 3 is the correct answer.
answered May 15, 2014 by meena.p
 
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