The equation of the directrix of the parabola | vedclass

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(C) The tangent at the vertex is given by $x-y+1=0 \dots (i)$.The focus of the parabola is $S(0,0)$.The distance from the focus to the tangent at the

Vedclass · Accepted Answer

$x-y+2=0$

Vedclass · Answer

(C) The tangent at the vertex is given by $x-y+1=0 \dots (i)$.The focus of the parabola is $S(0,0)$.The distance from the focus to the tangent at the vertex is $a = \left|\frac{0-0+1}{\sqrt{1^2+(-1)^2}}ight| = \frac{1}{\sqrt{2}}$.The directrix is parallel to the tangent at the vertex,so its equation is of the form $x-y+c=0$.The distance from the focus to the directrix is $2a = 2 	imes \frac{1}{\sqrt{2}} = \sqrt{2}$.Thus,$\left|\frac{0-0+c}{\sqrt{1^2+(-1)^2}}ight| = \sqrt{2}$ $\Rightarrow \frac{|c|}{\sqrt{2}} = \sqrt{2}$ $\Rightarrow |c| = 2$.Since the focus $(0,0)$ and the tangent at the vertex $x-y+1=0$ lie on the same side of the directrix,we choose $c=2$.Therefore,the equation of the directrix is $x-y+2=0$.

The equation of the directrix of the parabola whose focus is $(0,0)$ and the tangent at the vertex is $x-y+1=0$ is

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