Put the origin where the algebra is easiest, not where the picture suggests.
Find the receiver position
A satellite dish is 8 ft across and 2 ft deep at the centre. How far from the vertex should the receiver sit?
Put the vertex at the origin, opening upward: x² = 4cy. The rim is 4 ft out and 2 ft up, so the point (4, 2) is on the curve: 16 = 4c(2) = 8c, giving c = 2 ft. The receiver goes at the focus, 2 ft from the vertex. C uses the full 8 ft width as x instead of the half-width.
Nothing physical changes when you move the axes — only the amount of writing. Always take the free choice the problem is offering.
"6 ft across" gives x = 3, not 6. The same applies to an ellipse: a 40 m span means a = 20. This single step accounts for more wrong answers in conic word problems than every formula put together.
origin at the lowest point of the cable
so x² = 4cy
tower point: (40, 16)
half of 80, and 18 − 2
1600 = 4c(16), so 4c = 100
the curve is fixed
at x = 20: y = 400100 = 4
4 m above the lowest point
The answer the question wants is the height above the deck: 2 + 4 = 6 m. Working in "height above the lowest point" makes the algebra easy, but the final line has to translate back.
a = 25, b = 15
half the width, the full height
x²625 + y²225 = 1
centre at the base middle
400625 + y²225 = 1
put x = 20
y² = 225 × 925 = 81, y = 9 m
the clearance there
The width was halved and the height was not, because the arch is only the top half of the ellipse. Reading the picture decides that, not the formula.
Width and depth give one point on the curve; that point fixes c, and c is the focus distance.
A headlamp reflector is 24 cm across and 9 cm deep. Where does the bulb go?
x² = 4cy, vertex at the origin
the reflector opens upward
rim point (12, 9)
HALF of 24 across, the full 9 deep
144 = 4c(9) = 36c
substitute the rim point
c = 4 cm from the vertex
the bulb sits at the focus
Check: 4c = 16, and at x = 12, y = 14416 = 9 cm, the depth we were given.
With (24, 9) instead of (12, 9): 576 = 36c gives c = 16 cm, four times too far out. The width is squared, so forgetting to halve it multiplies the answer by 4, not 2.
The same fit-then-substitute method, three more ways.
An irrational answer: a semi-elliptical arch 12 m wide and 4 m high, 4 m from the centre
x²36 + y²16 = 1
a = 6, b = 4
1636 + y²16 = 1, so y²16 = 59
put x = 4
y² = 809, y = 4√53 ≈ 3.0 m
round only at the end
Not every height comes out whole. Keep the exact root until the last line.
Backwards: a dish 16 ft across has its receiver 3 ft from the vertex. How deep is it?
c = 3, so x² = 12y
the focus distance is given
rim x = 8: 64 = 12y
half of 16
y = 163 ≈ 5.3 ft deep
the unknown is now the depth
Here c is known and a coordinate is the unknown; the equation is the same.
Opening downward: a parabolic arch 12 m wide and 9 m high, 4 m from the centre
origin at the top: x² = −4cy
the arch opens down
foot of the arch (6, −9): 36 = 36c, c = 1
so x² = −4y
at x = 4: y = −4
4 m below the top
height = 9 − 4 = 5 m
translate back to the ground
With the origin at the top, points on the arch have negative y. The last line converts to height above the ground.
Every one of these problems ends by locating a focus or a point on the curve. Once the equation is fitted the rest is substitution — the hard part was choosing the axes and halving the width.
Tests describe a dish, reflector, cable or arch and ask for the focus distance or a height at some distance from the centre. The fastest route: put the origin at the vertex or centre, halve the width, substitute the one known point to fix the curve, then substitute again. A 10-second check: put your answer back as a point and see that it satisfies the fitted equation.
Find the focus distance
A parabolic reflector is 12 cm across and 3 cm deep. How far from the vertex is the focus?
With the vertex at the origin, the rim point is (6, 3) — half of 12, and the full depth. Then 36 = 4c(3) = 12c, so c = 3 cm. D uses the full 12 cm width as x, which quadruples the left side and doubles the answer.
Hint: the x-coordinate of the rim is half the width. The depth is measured from the vertex, so it is not halved.
