← Back to the Parabola Focus Directrix And Tangent Lines worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: G-GPE.A.2, G-C.A.4 · Level: Grade 10 Geometry
Problem 1 Focus and Directrix Distances
(a) Distance to focus: 6. Distance to the directrix: 3 − (−3) = 6. Equal, so on the parabola.
(b) Distance to focus: √(4² + 1²) = √17 ≈ 4.12. Distance to directrix: 2 + 3 = 5. Not equal.
(a) on the parabola (b) not on the parabola
Problem 2 Equation with Focus Above
The vertex is halfway between the focus and directrix, at (0, 0), and p = 6, so x² = 4(6)y.
x² = 24y
Problem 3 Equation Opening Downward
The vertex is (0, 0) and the focus is below it, so p = −7 and x² = 4(−7)y.
x² = −28y
Problem 4 Read the Focus
4p = 36, so p = 9. The focus is (0, 9) and the directrix is y = −9.
Focus (0, 9); directrix y = −9
Problem 5 Parabola Opening Right
The y is squared and the coefficient is positive, so it opens right. 4p = 24, so p = 6. The focus is (6, 0) and the directrix is x = −6.
Opens right; focus (6, 0); directrix x = −6
Problem 6 Vertex Form from Focus and Directrix
The vertex is halfway between: (2, 3). The focus is 2 above the vertex, so p = 2.
(x − 2)² = 4(2)(y − 3).
(x − 2)² = 8(y − 3)
Problem 7 Read a Shifted Parabola
The vertex is (−1, 4). 4p = −12, so p = −3 and the parabola opens down.
Focus: (−1, 4 − 3) = (−1, 1). Directrix: y = 4 + 3 = 7.
Vertex (−1, 4); focus (−1, 1); directrix y = 7
Problem 8 Equation Opening Right
The vertex is halfway between the focus and the directrix on the line y = −2: x = 3 + (−1)2 = 1, so the vertex is (1, −2). p = 3 − 1 = 2.
(y + 2)² = 4(2)(x − 1).
(y + 2)² = 8(x − 1)
Problem 9 Derive the Equation
For a point (x, y): √(x² + (y − 3)²) = y + 3.
Square both sides: x² + y² − 6y + 9 = y² + 6y + 9.
Cancel y² and 9: x² = 12y.
x² = 12y
Problem 10 A Point on the Parabola
36 = 8y, so y = 4.5.
The focus is (0, 2): distance = √(6² + 2.5²) = √42.25 = 6.5. The directrix is y = −2: distance = 4.5 + 2 = 6.5.
y = 4.5; both distances are 6.5
Problem 11 Width at the Focus
4p = 20, so p = 5 and the focus is (0, 5).
At y = 5: x² = 100, so x = ±10. The width is 20.
Focus (0, 5); width 20 (equal to 4p)
Problem 12 Complete the Square, Opening Up
Isolate the x-terms: x² − 6x = 8y − 17. Add 9: (x − 3)² = 8y − 8 = 8(y − 1).
Vertex (3, 1), 4p = 8 so p = 2. Focus (3, 3). Directrix y = −1.
Vertex (3, 1); focus (3, 3); directrix y = −1
Problem 13 Complete the Square, Opening Right
y² + 4y = 12x − 28. Add 4: (y + 2)² = 12x − 24 = 12(x − 2).
Vertex (2, −2), p = 3. Focus (5, −2). Directrix x = −1.
Vertex (2, −2); focus (5, −2); directrix x = −1
Problem 14 Congruent Tangent Segments
Tangent segments from one point are congruent: 6x − 5 = 4x + 13, so 2x = 18 and x = 9.
PA = 6(9) − 5 = 49. Check: PB = 4(9) + 13 = 49.
x = 9; PA = PB = 49
Problem 15 Rope to the Pond
The rope is tangent, so OT ⊥ PT and triangle OTP is right with hypotenuse 25.
PT = √(25² − 7²) = √576 = 24.
24 ft
Problem 16 Is the Line Tangent?
(a) 9² + 40² = 81 + 1,600 = 1,681 = 41². The converse of the Pythagorean Theorem gives a right angle at T: tangent.
(b) 8² + 14² = 64 + 196 = 260, but 16² = 256. Not equal, so no right angle: not tangent.
(a) tangent (b) not tangent
Problem 17 Satellite Dish Focus
The edge of the dish is at (12, 4.5). Use x² = 4py: 144 = 4p(4.5) = 18p, so p = 8.
The receiver sits 8 feet above the vertex.
Problem 18 Headlight Reflector
(a) 4p = 12, so p = 3. The bulb is 3 in above the vertex.
(b) At the edge x = 8: y = 6412 = 163 ≈ 5.33.
(a) 3 in (b) 16/3 ≈ 5.33 in deep
Problem 19 Construct and Justify
(a) ∠OTP is inscribed in a semicircle (its sides pass through the ends of the diameter OP), so ∠OTP = 90°. Then PT is perpendicular to the radius OT at T, which makes it tangent.
(b) PT = √(11² − 6²) = √85 ≈ 9.22.
(b) PT ≈ 9.22 cm
Problem 20 Angle Between Two Tangents
The line from P to the center bisects the 70° angle, so the angle at P in the right triangle is 35°.
OP = 8sin 35° ≈ 13.95. PT = 8tan 35° ≈ 11.43.
OP ≈ 13.95 in; PT ≈ 11.43 in