12 multiple-choice questions, progressively harder.
A parabola has focus (−4,6)(-4,6)(−4,6) and directrix y=−2y=-2y=−2. Which equation describes it?
Solution
Correct answer: C
The vertex is (−4,2)(-4,2)(−4,2), halfway from the focus to (−4,−2)(-4,-2)(−4,−2). The signed step to the focus is 4.
The axis is vertical and the opening is upward.
A parabola has focus (0,−5)(0,-5)(0,−5) and directrix x=6x=6x=6. Which equation describes it?
Correct answer: A
The vertex is (3,−5)(3,-5)(3,−5). The focus is 3 units left of it, so the signed parameter is −3-3−3.
What is the directrix of (x+7)2=−18(y−3)(x+7)^2=-18(y-3)(x+7)2=−18(y−3)?
Correct answer: D
The vertex is (−7,3)(-7,3)(−7,3) and p=−9/2p=-9/2p=−9/2. The directrix lies above the vertex.
What is the focus of y2+4y−20x+24=0y^2+4y-20x+24=0y2+4y−20x+24=0?
Correct answer: B
Complete the square.
The vertex is (1,−2)(1,-2)(1,−2) and p=5p=5p=5, so the focus is (6,−2)(6,-2)(6,−2).
A parabola has vertex (−2,5)(-2,5)(−2,5) and focus (−2,2)(-2,2)(−2,2). What is its equation?
The focus is below the vertex on a vertical axis. The signed step is 2−5=−32-5=-32−5=−3.
Use the x-gap as the squared quantity.
The focus is (1,−1)(1,-1)(1,−1) and the directrix is y=5y=5y=5.
At x=5x=5x=5, which height lies on the parabola?
The vertex is (1,2)(1,2)(1,2) and the signed parameter is −3-3−3.
Substitute x=5x=5x=5 into the vertical equation.
Which is the perpendicular foot from PPP at (−3,7)(-3,7)(−3,7) to the directrix x=2x=2x=2, and what is the line distance?
The perpendicular to a vertical line is horizontal, so keep the height 7. The foot must have x-coordinate 2.
The parabola (x−2)2=12(y+4)(x-2)^2=12(y+4)(x−2)2=12(y+4) moves 555 units left and 222 units up. What is its new equation?
The vertex moves from (2,−4)(2,-4)(2,−4) to (−3,−2)(-3,-2)(−3,−2). Translation keeps the signed step p=3p=3p=3.
A parabola has vertex (4,1)(4,1)(4,1) and focus (2,1)(2,1)(2,1). What is the focus-directrix spacing?
The vertex-focus distance is 2. The vertex is halfway between the focus and its directrix foot.
For focus (0,4)(0,4)(0,4) and directrix y=0y=0y=0, which equation comes directly from equating squared distances?
The squared distance to the focus uses (y−4)2(y-4)^2(y−4)2. The perpendicular distance to y=0y=0y=0 is ∣y∣|y|∣y∣.
Equate this with the squared line distance y2y^2y2.
Why can the squared distance equality be reversed to recover the original parabola locus?
If two nonnegative numbers have equal squares, they are equal. Here both quantities are distances.
The absolute value handles points on either side of the line.
A parabola has equation (y+6)2=8(x−1)(y+6)^2=8(x-1)(y+6)2=8(x−1).
At height y=−2y=-2y=−2, what is the focus distance of its point?
Substitution gives 16=8(x−1)16=8(x-1)16=8(x−1), so the point is (3,−2)(3,-2)(3,−2). The focus is (3,−6)(3,-6)(3,−6).
Its directrix is x=−1x=-1x=−1, also four units away.
Reset this practice set?
This clears every answer you have given and starts the set again from question 1.