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Conic sections questions
Parabola, ellipse and hyperbola are one family, the points whose distance from a focus is a fixed multiple of their distance from a line. That multiple, the eccentricity, determines which curve you get.
7 original questions · 27 marks · the conic sections notes · Further Pure 1
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Write down parametric coordinates for a general point on the parabola y² = 4ax and on the rectangular hyperbola xy = c².
Worked answer
Parabola: (at², 2at). Rectangular hyperbola: (ct, c/t). B1 B1 for the two parametrisations. Both turn a two-variable problem into algebra in a single parameter, and almost every tangent and normal question on this topic starts by using them.For the parabola y² = 20x, state a, the focus and the directrix.
Worked answer
4a = 20, so a = 5. The focus is (5, 0) and the directrix is x = −5. B1 for a = 5, B1 for the focus, B1 for the directrix. Quartering rather than halving the coefficient is the step most often slipped.For the ellipse x²/16 + y²/7 = 1, find the eccentricity, the foci and the equations of the directrices.
Worked answer
a = 4 and b² = 7. From b² = a²(1 − e²): 7 = 16(1 − e²), so e² = 9/16 and e = 3/4. Foci (±ae, 0) = (±3, 0); directrices x = ±a/e = ±16/3. M1 for using b² = a²(1 − e²), A1 for e = 3/4, A1 for the foci, A1 for the directrices. The minus sign inside the bracket is what distinguishes the ellipse formula from the hyperbola's.For the hyperbola x²/16 − y²/9 = 1, find the eccentricity and the foci, and state the difference of focal distances at the vertex (4, 0).
Worked answer
Here b² = a²(e² − 1): 9 = 16(e² − 1), so e² = 25/16 and e = 5/4. Foci (±5, 0). From (4, 0) the distances are 1 and 9, differing by 8 = 2a. M1 for using b² = a²(e² − 1), A1 for e = 5/4, A1 for the foci, B1 for the difference 2a. A hyperbola holds the difference of focal distances constant, as an ellipse holds the sum.State the focus-directrix property of a conic, and explain how the eccentricity decides which of the three curves appears.
Worked answer
Every point of the conic satisfies distance to the focus = e × distance to the directrix. When e < 1 the curve closes into an ellipse; e = 1 balances it into a parabola; e > 1 opens it into a hyperbola's two branches. B1 for the focus-directrix property, B1 B1 for the three cases of e. One definition with a dial produces all three.The point P(16, 16) lies on the parabola y² = 16x. Verify that it does, and check the focus-directrix property at P.
Worked answer
16² = 256 and 16 × 16 = 256, so P is on the curve. Here 4a = 16 and a = 4, so the focus is (4, 0) and the directrix is x = −4. The distance from P to the focus is √(12² + 16²) = 20, and the distance to the directrix is 16 + 4 = 20. Equal, exactly as e = 1 demands. B1 for the verification, B1 for the focus and directrix, M1 for a distance, A1 for both distances equal to 20.The parabola y² = 16x has focus S and directrix d. The tangent at the point P(16, 16) meets d at Q. Find the coordinates of Q and prove that angle PSQ is a right angle.
Worked answer
Here a = 4, so S is (4, 0) and d is x = −4. In parametric form P = (at², 2at) gives 2at = 16 with a = 4, so t = 2 and at² = 16 as required. Differentiating y² = 16x implicitly, 2y(dy/dx) = 16, so at P the gradient is 16/32 = 1/2 and the tangent is y − 16 = ½(x − 16), that is 2y = x + 16. Setting x = −4 gives y = 6, so Q(−4, 6). Then SP = (12, 16) and SQ = (−8, 6), and their scalar product is −96 + 96 = 0, so angle PSQ = 90°. B1 for S and d, M1 for differentiating to find the gradient at P, A1 for the tangent 2y = x + 16, M1 for putting x = −4, A1 for Q(−4, 6), M1 for the scalar product of SP and SQ, A1 for the conclusion. This holds for every point of every parabola, not only for P; repeating the working with (at², 2at) gives SP·SQ = 0 in general.
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