Derive the Equation of a Parabola - Geometry
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Identify the vertex for the parabola $y^2=-24x$.
Identify the vertex for the parabola $y^2=-24x$.
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$(0,0)$. Standard form $y^2=-24x$ has vertex at origin.
$(0,0)$. Standard form $y^2=-24x$ has vertex at origin.
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Identify the axis of symmetry for focus $(3,-2)$ and directrix $x=-1$.
Identify the axis of symmetry for focus $(3,-2)$ and directrix $x=-1$.
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$y=-2$. Horizontal line through focus for horizontal-axis parabola.
$y=-2$. Horizontal line through focus for horizontal-axis parabola.
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Identify the axis of symmetry for focus $(2,5)$ and directrix $y=1$.
Identify the axis of symmetry for focus $(2,5)$ and directrix $y=1$.
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$x=2$. Vertical line through focus for vertical-axis parabola.
$x=2$. Vertical line through focus for vertical-axis parabola.
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Find $p$ for a parabola with vertex $(4,-1)$ and focus $(1,-1)$.
Find $p$ for a parabola with vertex $(4,-1)$ and focus $(1,-1)$.
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$p=-3$. Distance from vertex to focus along negative $x$-direction.
$p=-3$. Distance from vertex to focus along negative $x$-direction.
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Identify the directrix for the parabola $x^2=16y$.
Identify the directrix for the parabola $x^2=16y$.
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$y=-4$. From $x^2=16y$, directrix is $p=4$ units below vertex.
$y=-4$. From $x^2=16y$, directrix is $p=4$ units below vertex.
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What equation results from the focus-directrix definition using squared distances?
What equation results from the focus-directrix definition using squared distances?
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$PF^2=\text{dist}(P,d)^2$. Squaring both sides eliminates absolute values in distance equation.
$PF^2=\text{dist}(P,d)^2$. Squaring both sides eliminates absolute values in distance equation.
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Identify the focus for the parabola $x^2=16y$.
Identify the focus for the parabola $x^2=16y$.
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$(0,4)$. From $x^2=16y$, we have $4p=16$, so $p=4$ and focus at $(0,4)$.
$(0,4)$. From $x^2=16y$, we have $4p=16$, so $p=4$ and focus at $(0,4)$.
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Identify the vertex for the parabola $x^2=16y$.
Identify the vertex for the parabola $x^2=16y$.
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$(0,0)$. Standard form $x^2=16y$ has vertex at origin.
$(0,0)$. Standard form $x^2=16y$ has vertex at origin.
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Find the focus for the parabola $(y+2)^2=-12(x+5)$.
Find the focus for the parabola $(y+2)^2=-12(x+5)$.
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$(-8,-2)$. From $(y+2)^2=-12(x+5)$, vertex $(-5)$ plus $p=-3$ gives $-8$.
$(-8,-2)$. From $(y+2)^2=-12(x+5)$, vertex $(-5)$ plus $p=-3$ gives $-8$.
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Find the directrix for the parabola $(y-4)^2=20(x-1)$.
Find the directrix for the parabola $(y-4)^2=20(x-1)$.
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$x=-4$. From $(y-4)^2=20(x-1)$, vertex $(1)$ minus $p=5$ gives $-4$.
$x=-4$. From $(y-4)^2=20(x-1)$, vertex $(1)$ minus $p=5$ gives $-4$.
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Find the focus for the parabola $(x+1)^2=-8(y-3)$.
Find the focus for the parabola $(x+1)^2=-8(y-3)$.
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$(-1,1)$. From $(x+1)^2=-8(y-3)$, vertex $(3)$ plus $p=-2$ gives $1$.
$(-1,1)$. From $(x+1)^2=-8(y-3)$, vertex $(3)$ plus $p=-2$ gives $1$.
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Find the directrix for the parabola $(x-2)^2=12(y+1)$.
Find the directrix for the parabola $(x-2)^2=12(y+1)$.
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$y=-4$. From $(x-2)^2=12(y+1)$, vertex $(-1)$ minus $p=3$ gives $-4$.
$y=-4$. From $(x-2)^2=12(y+1)$, vertex $(-1)$ minus $p=3$ gives $-4$.
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Derive the equation for focus $(2,-1)$ and directrix $x=6$.
Derive the equation for focus $(2,-1)$ and directrix $x=6$.
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$(y+1)^2=-8(x-4)$. Vertex at $(4,-1)$ with $p=-2$, giving $4p=-8$.
$(y+1)^2=-8(x-4)$. Vertex at $(4,-1)$ with $p=-2$, giving $4p=-8$.
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Derive the equation for focus $(5,6)$ and directrix $y=2$.
Derive the equation for focus $(5,6)$ and directrix $y=2$.
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$(x-5)^2=8(y-4)$. Vertex at $(5,4)$ with $p=2$, giving $4p=8$.
$(x-5)^2=8(y-4)$. Vertex at $(5,4)$ with $p=2$, giving $4p=8$.
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Derive the equation for focus $(-3,4)$ and directrix $x=1$.
Derive the equation for focus $(-3,4)$ and directrix $x=1$.
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$(y-4)^2=-16(x+1)$. Vertex at $(-1,4)$ with $p=-4$, giving $4p=-16$.
$(y-4)^2=-16(x+1)$. Vertex at $(-1,4)$ with $p=-4$, giving $4p=-16$.
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Derive the equation for focus $(1,2)$ and directrix $y=-2$.
Derive the equation for focus $(1,2)$ and directrix $y=-2$.
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$(x-1)^2=8(y-0)$. Vertex at $(1,0)$ with $p=2$, giving $4p=8$.
$(x-1)^2=8(y-0)$. Vertex at $(1,0)$ with $p=2$, giving $4p=8$.
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Find the equation of the parabola with focus $(4,2)$ and directrix $x=0$.
Find the equation of the parabola with focus $(4,2)$ and directrix $x=0$.
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$(y-2)^2=8(x-2)$. Vertex at $(2,2)$ with $p=2$, giving $4p=8$.
$(y-2)^2=8(x-2)$. Vertex at $(2,2)$ with $p=2$, giving $4p=8$.
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What is the distance formula from $P(x,y)$ to a vertical line $x=h$?
What is the distance formula from $P(x,y)$ to a vertical line $x=h$?
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$|x-h|$. Absolute value of difference between $x$-coordinate and line's $x$-value.
$|x-h|$. Absolute value of difference between $x$-coordinate and line's $x$-value.
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What is the distance formula from $P(x,y)$ to a horizontal line $y=k$?
What is the distance formula from $P(x,y)$ to a horizontal line $y=k$?
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$|y-k|$. Absolute value of difference between $y$-coordinate and line's $y$-value.
$|y-k|$. Absolute value of difference between $y$-coordinate and line's $y$-value.
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What is the defining property of a parabola using a focus $F$ and directrix line $d$?
What is the defining property of a parabola using a focus $F$ and directrix line $d$?
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$PF=\text{dist}(P,d)$ for all points $P$ on the parabola. Each point on the parabola is equidistant from focus and directrix.
$PF=\text{dist}(P,d)$ for all points $P$ on the parabola. Each point on the parabola is equidistant from focus and directrix.
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Identify the directrix for the parabola $y^2=-24x$.
Identify the directrix for the parabola $y^2=-24x$.
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$x=6$. From $y^2=-24x$, directrix is $|p|=6$ units right of vertex.
$x=6$. From $y^2=-24x$, directrix is $|p|=6$ units right of vertex.
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Derive the equation from focus $(0,2)$ and directrix $y=-2$ using $PF=\text{dist}(P,d)$.
Derive the equation from focus $(0,2)$ and directrix $y=-2$ using $PF=\text{dist}(P,d)$.
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$x^2=8y$. Setting $PF=\text{dist}(P,d)$ and squaring gives vertex at origin with $4p=8$.
$x^2=8y$. Setting $PF=\text{dist}(P,d)$ and squaring gives vertex at origin with $4p=8$.
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Derive the equation from focus $(2,0)$ and directrix $x=-2$ using $PF=\text{dist}(P,d)$.
Derive the equation from focus $(2,0)$ and directrix $x=-2$ using $PF=\text{dist}(P,d)$.
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$y^2=8x$. Setting $PF=\text{dist}(P,d)$ and squaring gives vertex at origin with $4p=8$.
$y^2=8x$. Setting $PF=\text{dist}(P,d)$ and squaring gives vertex at origin with $4p=8$.
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Find the equation of the parabola with focus $(0,-1)$ and directrix $y=3$.
Find the equation of the parabola with focus $(0,-1)$ and directrix $y=3$.
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$x^2=-8(y-1)$. Vertex at $(0,1)$ with $p=-2$, giving $4p=-8$.
$x^2=-8(y-1)$. Vertex at $(0,1)$ with $p=-2$, giving $4p=-8$.
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What is the distance formula from $P(x,y)$ to a line $Ax+By+C=0$?
What is the distance formula from $P(x,y)$ to a line $Ax+By+C=0$?
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$\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$. Standard point-to-line distance formula using coefficients and normal vector.
$\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$. Standard point-to-line distance formula using coefficients and normal vector.
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Identify the focus for the parabola $y^2=-24x$.
Identify the focus for the parabola $y^2=-24x$.
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$(-6,0)$. From $y^2=-24x$, we have $4p=-24$, so $p=-6$ and focus at $(-6,0)$.
$(-6,0)$. From $y^2=-24x$, we have $4p=-24$, so $p=-6$ and focus at $(-6,0)$.
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Derive the equation for focus $ (0,-5) $ and directrix $ y=1 $.
Derive the equation for focus $ (0,-5) $ and directrix $ y=1 $.
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$x^2=-12(y+2)$. Vertex at $ (0,-2) $ with $ p=-3 $, giving $ 4p=-12 $.
$x^2=-12(y+2)$. Vertex at $ (0,-2) $ with $ p=-3 $, giving $ 4p=-12 $.
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Find $p$ for a parabola with vertex $(0,2)$ and focus $(0,5)$.
Find $p$ for a parabola with vertex $(0,2)$ and focus $(0,5)$.
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$p=3$. Distance from vertex to focus along positive $y$-direction.
$p=3$. Distance from vertex to focus along positive $y$-direction.
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Derive the equation for focus $(0,0)$ and directrix $x=4$.
Derive the equation for focus $(0,0)$ and directrix $x=4$.
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$y^2=-8(x-2)$. Vertex at $(2,0)$ with $p=-2$, giving $4p=-8$.
$y^2=-8(x-2)$. Vertex at $(2,0)$ with $p=-2$, giving $4p=-8$.
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Derive the equation for focus $(0,0)$ and directrix $y=4$.
Derive the equation for focus $(0,0)$ and directrix $y=4$.
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$x^2=-8(y-2)$. Vertex at $(0,2)$ with $p=-2$, giving $4p=-8$.
$x^2=-8(y-2)$. Vertex at $(0,2)$ with $p=-2$, giving $4p=-8$.
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