Find the equation of the parabola whose focus is \left(2{,}2\right) and directrix is y=4.
A parabola is the set of all points that are equidistant to the focus and the directrix.
The distance of the directrix and the parabola is y-4.
The distance of the parabola and the focus can be obtained using the distance formula:
\sqrt{\left(x-2\right)^2+\left(y-2\right)^2}
Therefore an equation is:
\left(y-4\right)^2 = \left(x-2\right)^2+ \left(y-2\right)^2
y^2-8y+16 = x^2 -4x+4 + y^2-4y+4
-4y = x^2 -4x-8
y= -\dfrac{1}{4}x^2+x+2
Find the equation of the parabola whose focus is \left(-1{,}3\right) and directrix is y=1.
A parabola is the set of all points that are equidistant to the focus and the directrix.
The distance of the directrix and the parabola is y-1.
The distance of the parabola and the focus can be obtained using the distance formula:
\sqrt{\left(x+1\right)^2+\left(y-3\right)^2}
Therefore an equation is:
\left(y-1\right)^2 = \left(x+1\right)^2+ \left(y-3\right)^2
y^2-2y+1 = x^2 +2x+1 + y^2-6y+9
4y = x^2 +2x+9
y=\dfrac{1}{4}x^2+\dfrac{2}{4}x+\dfrac{9}{4}
y=\dfrac{1}{4}x^2+\dfrac{1}{2}x+\dfrac{9}{4}
y=\dfrac{1}{4}x^2+\dfrac{1}{2}x+\dfrac{9}{4}
Find the equation of the parabola whose focus is \left(0{,}2\right) and directrix is y=-1.
A parabola is the set of all points that are equidistant to the focus and the directrix.
The distance of the directrix and the parabola is y+1.
The distance of the parabola and the focus can be obtained using the distance formula:
\sqrt{\left(x-0\right)^2+\left(y-2\right)^2}
Therefore an equation is:
\left(y+1\right)^2 = \left(x\right)^2+ \left(y-2\right)^2
y^2+2y+1 = x^2 + y^2-4y+4
6y = x^2+3
y=\dfrac{1}{6}x^2+\dfrac{3}{6}
y=\dfrac{1}{6}x^2+\dfrac{1}{2}
y=\dfrac{1}{6}x^2+\dfrac{1}{2}
Find the equation of the parabola whose focus is \left(2{,}0\right) and directrix is y=3.
A parabola is the set of all points that are equidistant to the focus and the directrix.
The distance of the directrix and the parabola is y-3.
The distance of the parabola and the focus can be obtained using the distance formula:
\sqrt{\left(x-2\right)^2+\left(y-0\right)^2}
Therefore an equation is:
\left(y-3\right)^2 = \left(x-2\right)^2+ y^2
y^2-6y+9=x^2-4x+4+y^2
-6y = x^2 -4x-5
y=-\dfrac{1}{6}x^2+\dfrac{4}{6}x+\dfrac{5}{6}
y=-\dfrac{1}{6}x^2+\dfrac{2}{3}x+\dfrac{5}{6}
y=-\dfrac{1}{6}x^2+\dfrac{2}{3}x+\dfrac{5}{6}
Find the equation of the parabola whose focus is \left(-1{,}2\right) and directrix is x=1.
A parabola is the set of all points that are equidistant to the focus and the directrix.
The distance of the directrix and the parabola is x-1.
The distance of the parabola and the focus can be obtained using the distance formula:
\sqrt{\left(x+1\right)^2+\left(y-2\right)^2}
Therefore an equation is:
\left(x-1\right)^2 = \left(x+1\right)^2+ \left(y-2\right)^2
x^2-2x+1=x^2+2x+1+y^2-4y+4
-4x = y^2 -4y+4
x=-\dfrac{1}{4}y^2+y-1
x=-\dfrac{1}{4}y^2+y-1
Find the equation of the parabola whose focus is \left(1{,}4\right) and directrix is x=-1.
A parabola is the set of all points that are equidistant to the focus and the directrix.
The distance of the directrix and the parabola is x+1.
The distance of the parabola and the focus can be obtained using the distance formula:
\sqrt{\left(x-1\right)^2+\left(y-4\right)^2}
Therefore an equation is:
\left(x+1\right)^2 = \left(x-1\right)^2+ \left(y-4\right)^2
x^2+2x+1 = x^2-2x+1+y^2-8y+16
4x=y^2-8y+16
x=\dfrac{1}{4}y^2-\dfrac{8}{4}y+\dfrac{16}{4}
x=\dfrac{1}{4}y^2-2y+4
x=\dfrac{1}{4}y^2-2y+4
Find the equation of the parabola whose focus is \left(2,-1\right) and directrix is x=3.
A parabola is the set of all points that are equidistant to the focus and the directrix.
The distance of the directrix and the parabola is x-3.
The distance of the parabola and the focus can be obtained using the distance formula:
\sqrt{\left(x-2\right)^2+\left(y+1\right)^2}
Therefore an equation is:
\left(x-3\right)^2 = \left(x-2\right)^2+ \left(y+1\right)^2
x^2-6x+9 = x^2 -4x+4 + y^2+2y+1
-2x =y^2 +2y-4
x=-\dfrac{1}{2}y^2-y+2
x= -\dfrac{1}{2}y^2-y+2