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directrix y=x^2+1

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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directrix y=x²+1

Solution

y=34

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Parabola Directrix

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (the directrix)

Parabola standard equation

4p(y−k)=(x−h)² is the standard equation for an up-down facing parabola with vertex at (h, k),
and a focal length |p|

Rewrite y=x²+1 in the standard form: 4· 14 (y−1)=(x−0)²

(h, k)=(0, 1), p=14

Parabola is symmetric around the y-axis and so the directrix is a line parallel to the x-axis, a distance −p from the center (0, 1) y-coordinate

y=1−p

y=1−14

Refine

y=34

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こんにちは, について質問する directrix y=x²+1 または以下を選択してください

−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

directrix y=x^2+1

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